Describe a list of numbers that has alternating functions without using trigonometric functions?
$begingroup$
The list is: $ {5,6,10,12,15,18,20,24,25...}$
Is there a way to describe this list of numbers mathematically without using $sin(x)
, cos(x), cos(nx)$, etc?
There is a way using the trigonometric function $sin$. Simply use the jumping between $0$ and $1$ of $$frac{1+sin(npi)}{2}$$ and $$frac{1+sinleft(npi+frac{3pi}{2}right)}{2 }$$ which instead oscillates between $1$ and $0$. Then you can get the answer by describing $A = {5, x, 10, y, 15, z, 20}$ as well as $ B = {x, 6, y, 12, z, 18}$. Every other number in the lists $A$ and $B$ will either be numbers or they will be zero, since they're multiplied with said $sin$ function. $A + B$ is therefore a list containing two different methods of sorting, i.e. $A$'s function could be $ x^2 $ and $B$'s function could be $x^3-x^2+2x $ where every item in the list $C$, where $C = A + B$, is based on a different equation than the one before and after that said item.
Now, again, is there a way to create this said list $C$, containing the pattern of two very different functions that create the list $C $ where every item in the slot $ 2n$ is connected to one function but that function is different from the item in the slot $2n + 1$?
sequences-and-series
$endgroup$
add a comment |
$begingroup$
The list is: $ {5,6,10,12,15,18,20,24,25...}$
Is there a way to describe this list of numbers mathematically without using $sin(x)
, cos(x), cos(nx)$, etc?
There is a way using the trigonometric function $sin$. Simply use the jumping between $0$ and $1$ of $$frac{1+sin(npi)}{2}$$ and $$frac{1+sinleft(npi+frac{3pi}{2}right)}{2 }$$ which instead oscillates between $1$ and $0$. Then you can get the answer by describing $A = {5, x, 10, y, 15, z, 20}$ as well as $ B = {x, 6, y, 12, z, 18}$. Every other number in the lists $A$ and $B$ will either be numbers or they will be zero, since they're multiplied with said $sin$ function. $A + B$ is therefore a list containing two different methods of sorting, i.e. $A$'s function could be $ x^2 $ and $B$'s function could be $x^3-x^2+2x $ where every item in the list $C$, where $C = A + B$, is based on a different equation than the one before and after that said item.
Now, again, is there a way to create this said list $C$, containing the pattern of two very different functions that create the list $C $ where every item in the slot $ 2n$ is connected to one function but that function is different from the item in the slot $2n + 1$?
sequences-and-series
$endgroup$
3
$begingroup$
Very simply: it's the sequence $(5n,6n)_{nge 1}$.
$endgroup$
– Bernard
Dec 24 '18 at 19:24
2
$begingroup$
$f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
$endgroup$
– Anurag A
Dec 24 '18 at 19:26
1
$begingroup$
To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
$endgroup$
– David C. Ullrich
Dec 24 '18 at 19:55
add a comment |
$begingroup$
The list is: $ {5,6,10,12,15,18,20,24,25...}$
Is there a way to describe this list of numbers mathematically without using $sin(x)
, cos(x), cos(nx)$, etc?
There is a way using the trigonometric function $sin$. Simply use the jumping between $0$ and $1$ of $$frac{1+sin(npi)}{2}$$ and $$frac{1+sinleft(npi+frac{3pi}{2}right)}{2 }$$ which instead oscillates between $1$ and $0$. Then you can get the answer by describing $A = {5, x, 10, y, 15, z, 20}$ as well as $ B = {x, 6, y, 12, z, 18}$. Every other number in the lists $A$ and $B$ will either be numbers or they will be zero, since they're multiplied with said $sin$ function. $A + B$ is therefore a list containing two different methods of sorting, i.e. $A$'s function could be $ x^2 $ and $B$'s function could be $x^3-x^2+2x $ where every item in the list $C$, where $C = A + B$, is based on a different equation than the one before and after that said item.
Now, again, is there a way to create this said list $C$, containing the pattern of two very different functions that create the list $C $ where every item in the slot $ 2n$ is connected to one function but that function is different from the item in the slot $2n + 1$?
sequences-and-series
$endgroup$
The list is: $ {5,6,10,12,15,18,20,24,25...}$
Is there a way to describe this list of numbers mathematically without using $sin(x)
, cos(x), cos(nx)$, etc?
There is a way using the trigonometric function $sin$. Simply use the jumping between $0$ and $1$ of $$frac{1+sin(npi)}{2}$$ and $$frac{1+sinleft(npi+frac{3pi}{2}right)}{2 }$$ which instead oscillates between $1$ and $0$. Then you can get the answer by describing $A = {5, x, 10, y, 15, z, 20}$ as well as $ B = {x, 6, y, 12, z, 18}$. Every other number in the lists $A$ and $B$ will either be numbers or they will be zero, since they're multiplied with said $sin$ function. $A + B$ is therefore a list containing two different methods of sorting, i.e. $A$'s function could be $ x^2 $ and $B$'s function could be $x^3-x^2+2x $ where every item in the list $C$, where $C = A + B$, is based on a different equation than the one before and after that said item.
Now, again, is there a way to create this said list $C$, containing the pattern of two very different functions that create the list $C $ where every item in the slot $ 2n$ is connected to one function but that function is different from the item in the slot $2n + 1$?
sequences-and-series
sequences-and-series
edited Dec 24 '18 at 23:37
Dave
8,78711033
8,78711033
asked Dec 24 '18 at 19:22
lmaolmao
83
83
3
$begingroup$
Very simply: it's the sequence $(5n,6n)_{nge 1}$.
$endgroup$
– Bernard
Dec 24 '18 at 19:24
2
$begingroup$
$f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
$endgroup$
– Anurag A
Dec 24 '18 at 19:26
1
$begingroup$
To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
$endgroup$
– David C. Ullrich
Dec 24 '18 at 19:55
add a comment |
3
$begingroup$
Very simply: it's the sequence $(5n,6n)_{nge 1}$.
$endgroup$
– Bernard
Dec 24 '18 at 19:24
2
$begingroup$
$f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
$endgroup$
– Anurag A
Dec 24 '18 at 19:26
1
$begingroup$
To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
$endgroup$
– David C. Ullrich
Dec 24 '18 at 19:55
3
3
$begingroup$
Very simply: it's the sequence $(5n,6n)_{nge 1}$.
$endgroup$
– Bernard
Dec 24 '18 at 19:24
$begingroup$
Very simply: it's the sequence $(5n,6n)_{nge 1}$.
$endgroup$
– Bernard
Dec 24 '18 at 19:24
2
2
$begingroup$
$f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
$endgroup$
– Anurag A
Dec 24 '18 at 19:26
$begingroup$
$f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
$endgroup$
– Anurag A
Dec 24 '18 at 19:26
1
1
$begingroup$
To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
$endgroup$
– David C. Ullrich
Dec 24 '18 at 19:55
$begingroup$
To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
$endgroup$
– David C. Ullrich
Dec 24 '18 at 19:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
What you have is three sequences
$$ a_1,a_2,a_3,cdots $$
$$ b_1,b_2,b_3,cdots$$
with a third sequence which is a 'shuffle' of the first two
$$ a_1,b_1,a_2,b_2,a_3,b_3,cdots=c_1,c_2,c_3,c_4,c_5,c_6cdots$$
Define two functions on $mathbb{N}$:
- $k(n)=leftlfloordfrac{n+1}{2}rightrfloor$
- $h(n)=dfrac{1+(-1)^{n+1}}{2}$
Then $k$ defines a sequence $1,1,2,2,3,3,cdots$ and
$h$ defines a sequence $1,0,1,0,1,0,cdots$
Then the sequence $c_n$ can be defined
$$ c_n=a_{k(n)}h(n)+b_{k(n)}h(n+1) $$
Then
begin{eqnarray}
c_1&=&a_{k(1)}h(1)+b_{k(1)}h(2)\
&=&a_1cdot1+b_1cdot0\
&=&a_1
end{eqnarray}
begin{eqnarray}
c_2&=&a_{k(2)}h(2)+b_{k(2)}h(3)\
&=&a_1cdot0+b_1cdot1\
&=&b_1
end{eqnarray}
etc.
Substituting in your particular sequences will result in
$$c_n=left(frac{11+(-1)^n}{2}right)k_n$$
$endgroup$
add a comment |
$begingroup$
If you just want to define a sequence so that the terms come from one function and then another, alternately, a simple way is:
$$
c_n = begin{cases}
f(n) & text{if $n equiv 0 pmod 2$}, \
g(n) & text{if $n equiv 1 pmod 2$}.
end{cases}
$$
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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oldest
votes
active
oldest
votes
$begingroup$
What you have is three sequences
$$ a_1,a_2,a_3,cdots $$
$$ b_1,b_2,b_3,cdots$$
with a third sequence which is a 'shuffle' of the first two
$$ a_1,b_1,a_2,b_2,a_3,b_3,cdots=c_1,c_2,c_3,c_4,c_5,c_6cdots$$
Define two functions on $mathbb{N}$:
- $k(n)=leftlfloordfrac{n+1}{2}rightrfloor$
- $h(n)=dfrac{1+(-1)^{n+1}}{2}$
Then $k$ defines a sequence $1,1,2,2,3,3,cdots$ and
$h$ defines a sequence $1,0,1,0,1,0,cdots$
Then the sequence $c_n$ can be defined
$$ c_n=a_{k(n)}h(n)+b_{k(n)}h(n+1) $$
Then
begin{eqnarray}
c_1&=&a_{k(1)}h(1)+b_{k(1)}h(2)\
&=&a_1cdot1+b_1cdot0\
&=&a_1
end{eqnarray}
begin{eqnarray}
c_2&=&a_{k(2)}h(2)+b_{k(2)}h(3)\
&=&a_1cdot0+b_1cdot1\
&=&b_1
end{eqnarray}
etc.
Substituting in your particular sequences will result in
$$c_n=left(frac{11+(-1)^n}{2}right)k_n$$
$endgroup$
add a comment |
$begingroup$
What you have is three sequences
$$ a_1,a_2,a_3,cdots $$
$$ b_1,b_2,b_3,cdots$$
with a third sequence which is a 'shuffle' of the first two
$$ a_1,b_1,a_2,b_2,a_3,b_3,cdots=c_1,c_2,c_3,c_4,c_5,c_6cdots$$
Define two functions on $mathbb{N}$:
- $k(n)=leftlfloordfrac{n+1}{2}rightrfloor$
- $h(n)=dfrac{1+(-1)^{n+1}}{2}$
Then $k$ defines a sequence $1,1,2,2,3,3,cdots$ and
$h$ defines a sequence $1,0,1,0,1,0,cdots$
Then the sequence $c_n$ can be defined
$$ c_n=a_{k(n)}h(n)+b_{k(n)}h(n+1) $$
Then
begin{eqnarray}
c_1&=&a_{k(1)}h(1)+b_{k(1)}h(2)\
&=&a_1cdot1+b_1cdot0\
&=&a_1
end{eqnarray}
begin{eqnarray}
c_2&=&a_{k(2)}h(2)+b_{k(2)}h(3)\
&=&a_1cdot0+b_1cdot1\
&=&b_1
end{eqnarray}
etc.
Substituting in your particular sequences will result in
$$c_n=left(frac{11+(-1)^n}{2}right)k_n$$
$endgroup$
add a comment |
$begingroup$
What you have is three sequences
$$ a_1,a_2,a_3,cdots $$
$$ b_1,b_2,b_3,cdots$$
with a third sequence which is a 'shuffle' of the first two
$$ a_1,b_1,a_2,b_2,a_3,b_3,cdots=c_1,c_2,c_3,c_4,c_5,c_6cdots$$
Define two functions on $mathbb{N}$:
- $k(n)=leftlfloordfrac{n+1}{2}rightrfloor$
- $h(n)=dfrac{1+(-1)^{n+1}}{2}$
Then $k$ defines a sequence $1,1,2,2,3,3,cdots$ and
$h$ defines a sequence $1,0,1,0,1,0,cdots$
Then the sequence $c_n$ can be defined
$$ c_n=a_{k(n)}h(n)+b_{k(n)}h(n+1) $$
Then
begin{eqnarray}
c_1&=&a_{k(1)}h(1)+b_{k(1)}h(2)\
&=&a_1cdot1+b_1cdot0\
&=&a_1
end{eqnarray}
begin{eqnarray}
c_2&=&a_{k(2)}h(2)+b_{k(2)}h(3)\
&=&a_1cdot0+b_1cdot1\
&=&b_1
end{eqnarray}
etc.
Substituting in your particular sequences will result in
$$c_n=left(frac{11+(-1)^n}{2}right)k_n$$
$endgroup$
What you have is three sequences
$$ a_1,a_2,a_3,cdots $$
$$ b_1,b_2,b_3,cdots$$
with a third sequence which is a 'shuffle' of the first two
$$ a_1,b_1,a_2,b_2,a_3,b_3,cdots=c_1,c_2,c_3,c_4,c_5,c_6cdots$$
Define two functions on $mathbb{N}$:
- $k(n)=leftlfloordfrac{n+1}{2}rightrfloor$
- $h(n)=dfrac{1+(-1)^{n+1}}{2}$
Then $k$ defines a sequence $1,1,2,2,3,3,cdots$ and
$h$ defines a sequence $1,0,1,0,1,0,cdots$
Then the sequence $c_n$ can be defined
$$ c_n=a_{k(n)}h(n)+b_{k(n)}h(n+1) $$
Then
begin{eqnarray}
c_1&=&a_{k(1)}h(1)+b_{k(1)}h(2)\
&=&a_1cdot1+b_1cdot0\
&=&a_1
end{eqnarray}
begin{eqnarray}
c_2&=&a_{k(2)}h(2)+b_{k(2)}h(3)\
&=&a_1cdot0+b_1cdot1\
&=&b_1
end{eqnarray}
etc.
Substituting in your particular sequences will result in
$$c_n=left(frac{11+(-1)^n}{2}right)k_n$$
edited Dec 24 '18 at 21:59
answered Dec 24 '18 at 21:47
John Wayland BalesJohn Wayland Bales
14.1k21238
14.1k21238
add a comment |
add a comment |
$begingroup$
If you just want to define a sequence so that the terms come from one function and then another, alternately, a simple way is:
$$
c_n = begin{cases}
f(n) & text{if $n equiv 0 pmod 2$}, \
g(n) & text{if $n equiv 1 pmod 2$}.
end{cases}
$$
$endgroup$
add a comment |
$begingroup$
If you just want to define a sequence so that the terms come from one function and then another, alternately, a simple way is:
$$
c_n = begin{cases}
f(n) & text{if $n equiv 0 pmod 2$}, \
g(n) & text{if $n equiv 1 pmod 2$}.
end{cases}
$$
$endgroup$
add a comment |
$begingroup$
If you just want to define a sequence so that the terms come from one function and then another, alternately, a simple way is:
$$
c_n = begin{cases}
f(n) & text{if $n equiv 0 pmod 2$}, \
g(n) & text{if $n equiv 1 pmod 2$}.
end{cases}
$$
$endgroup$
If you just want to define a sequence so that the terms come from one function and then another, alternately, a simple way is:
$$
c_n = begin{cases}
f(n) & text{if $n equiv 0 pmod 2$}, \
g(n) & text{if $n equiv 1 pmod 2$}.
end{cases}
$$
answered Dec 24 '18 at 22:03
David KDavid K
54k342116
54k342116
add a comment |
add a comment |
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3
$begingroup$
Very simply: it's the sequence $(5n,6n)_{nge 1}$.
$endgroup$
– Bernard
Dec 24 '18 at 19:24
2
$begingroup$
$f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
$endgroup$
– Anurag A
Dec 24 '18 at 19:26
1
$begingroup$
To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
$endgroup$
– David C. Ullrich
Dec 24 '18 at 19:55