Does this set of coordinates result in a curve?
$begingroup$
Coordinates: $(0,0), (3,3), (6,4.5), (9, 5.25)$
If this is a curve is there a formula for determining the $y$ value for any given $x$ within the range $0$ to $9?$
algebra-precalculus
edited Dec 10 '18 at 15:37
user376343
3,9584829
asked Jul 19 '16 at 3:27
TomTom
1
$endgroup$
add a comment |
$begingroup$
Coordinates: $(0,0), (3,3), (6,4.5), (9, 5.25)$
If this is a curve is there a formula for determining the $y$ value for any given $x$ within the range $0$ to $9?$
algebra-precalculus
edited Dec 10 '18 at 15:37
user376343
3,9584829
asked Jul 19 '16 at 3:27
TomTom
1
$endgroup$
1
$begingroup$
What type of curve are you talking about ?
$endgroup$
– Vivek Kaushik
Jul 19 '16 at 3:29
1
$begingroup$
Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
$endgroup$
– hardmath
Jul 19 '16 at 3:39
add a comment |
$begingroup$
Coordinates: $(0,0), (3,3), (6,4.5), (9, 5.25)$
If this is a curve is there a formula for determining the $y$ value for any given $x$ within the range $0$ to $9?$
algebra-precalculus
edited Dec 10 '18 at 15:37
user376343
3,9584829
asked Jul 19 '16 at 3:27
TomTom
1
$endgroup$
Coordinates: $(0,0), (3,3), (6,4.5), (9, 5.25)$
If this is a curve is there a formula for determining the $y$ value for any given $x$ within the range $0$ to $9?$
algebra-precalculus
algebra-precalculus
edited Dec 10 '18 at 15:37
user376343
3,9584829
asked Jul 19 '16 at 3:27
TomTom
1
edited Dec 10 '18 at 15:37
user376343
3,9584829
asked Jul 19 '16 at 3:27
TomTom
1
edited Dec 10 '18 at 15:37
user376343
3,9584829
edited Dec 10 '18 at 15:37
user376343
3,9584829
edited Dec 10 '18 at 15:37
user376343
3,9584829
3,9584829
asked Jul 19 '16 at 3:27
TomTom
1
asked Jul 19 '16 at 3:27
TomTom
1
asked Jul 19 '16 at 3:27
TomTom
1
1
1
$begingroup$
What type of curve are you talking about ?
$endgroup$
– Vivek Kaushik
Jul 19 '16 at 3:29
1
$begingroup$
Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
$endgroup$
– hardmath
Jul 19 '16 at 3:39
add a comment |
1
$begingroup$
What type of curve are you talking about ?
$endgroup$
– Vivek Kaushik
Jul 19 '16 at 3:29
1
$begingroup$
Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
$endgroup$
– hardmath
Jul 19 '16 at 3:39
1
1
$begingroup$
What type of curve are you talking about ?
$endgroup$
– Vivek Kaushik
Jul 19 '16 at 3:29
$begingroup$
What type of curve are you talking about ?
$endgroup$
– Vivek Kaushik
Jul 19 '16 at 3:29
1
1
$begingroup$
Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
$endgroup$
– hardmath
Jul 19 '16 at 3:39
$begingroup$
Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
$endgroup$
– hardmath
Jul 19 '16 at 3:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I recommend looking at Lagrange's Interpolation Formula. Given any $n$ points, they can always be interpolated by a polynomial of degree $n-1$ or less. This means that those points will always lie on the curve of that polynomial. So given your 4 points, there is a polynomial of degree 3 or less such that the points are on the curve of that polynomial. The explicit equation for the polynomial can be found using the formula included in the link.
However, these four points most definitely do not determine uniquely a continuous function (a curve). There are very many curves that could pass through these points.
answered Jul 19 '16 at 3:41
ChristianChristian
2,1111522
$endgroup$
1
$begingroup$
I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
$endgroup$
– Jean Marie
Jul 19 '16 at 7:03
add a comment |
$begingroup$
This particular curve can be also $f(x)=6(1-2^{-x/3})$. Here is how I guessed this function:
you go in steps of 3, so you might have something $x/3$.
At the first step the function increases by 3, then at the next step by 3/2, then 3/4. From here, I would guess that the formula is $$sum_s3left(frac{1}{2}right)^s,$$ where $s=x/3$
$sum_s(1/2)^s=frac{1-(1/2)^{s+1}}{1-1/2}$
answered Jul 19 '16 at 5:14
AndreiAndrei
13.2k21230
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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votes
active
oldest
votes
$begingroup$
I recommend looking at Lagrange's Interpolation Formula. Given any $n$ points, they can always be interpolated by a polynomial of degree $n-1$ or less. This means that those points will always lie on the curve of that polynomial. So given your 4 points, there is a polynomial of degree 3 or less such that the points are on the curve of that polynomial. The explicit equation for the polynomial can be found using the formula included in the link.
However, these four points most definitely do not determine uniquely a continuous function (a curve). There are very many curves that could pass through these points.
answered Jul 19 '16 at 3:41
ChristianChristian
2,1111522
$endgroup$
1
$begingroup$
I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
$endgroup$
– Jean Marie
Jul 19 '16 at 7:03
add a comment |
$begingroup$
I recommend looking at Lagrange's Interpolation Formula. Given any $n$ points, they can always be interpolated by a polynomial of degree $n-1$ or less. This means that those points will always lie on the curve of that polynomial. So given your 4 points, there is a polynomial of degree 3 or less such that the points are on the curve of that polynomial. The explicit equation for the polynomial can be found using the formula included in the link.
However, these four points most definitely do not determine uniquely a continuous function (a curve). There are very many curves that could pass through these points.
answered Jul 19 '16 at 3:41
ChristianChristian
2,1111522
$endgroup$
1
$begingroup$
I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
$endgroup$
– Jean Marie
Jul 19 '16 at 7:03
add a comment |
$begingroup$
I recommend looking at Lagrange's Interpolation Formula. Given any $n$ points, they can always be interpolated by a polynomial of degree $n-1$ or less. This means that those points will always lie on the curve of that polynomial. So given your 4 points, there is a polynomial of degree 3 or less such that the points are on the curve of that polynomial. The explicit equation for the polynomial can be found using the formula included in the link.
However, these four points most definitely do not determine uniquely a continuous function (a curve). There are very many curves that could pass through these points.
answered Jul 19 '16 at 3:41
ChristianChristian
2,1111522
$endgroup$
I recommend looking at Lagrange's Interpolation Formula. Given any $n$ points, they can always be interpolated by a polynomial of degree $n-1$ or less. This means that those points will always lie on the curve of that polynomial. So given your 4 points, there is a polynomial of degree 3 or less such that the points are on the curve of that polynomial. The explicit equation for the polynomial can be found using the formula included in the link.
However, these four points most definitely do not determine uniquely a continuous function (a curve). There are very many curves that could pass through these points.
answered Jul 19 '16 at 3:41
ChristianChristian
2,1111522
answered Jul 19 '16 at 3:41
ChristianChristian
2,1111522
answered Jul 19 '16 at 3:41
ChristianChristian
2,1111522
answered Jul 19 '16 at 3:41
ChristianChristian
2,1111522
2,1111522
1
$begingroup$
I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
$endgroup$
– Jean Marie
Jul 19 '16 at 7:03
add a comment |
1
$begingroup$
I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
$endgroup$
– Jean Marie
Jul 19 '16 at 7:03
1
1
$begingroup$
I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
$endgroup$
– Jean Marie
Jul 19 '16 at 7:03
$begingroup$
I agree with Christian. This Lagrange interpolation formula will give you the polynomial $$f(x)=dfrac{4}{3}x-dfrac{1}{8}x^2+dfrac{1}{216}x^3$$
$endgroup$
– Jean Marie
Jul 19 '16 at 7:03
add a comment |
$begingroup$
This particular curve can be also $f(x)=6(1-2^{-x/3})$. Here is how I guessed this function:
you go in steps of 3, so you might have something $x/3$.
At the first step the function increases by 3, then at the next step by 3/2, then 3/4. From here, I would guess that the formula is $$sum_s3left(frac{1}{2}right)^s,$$ where $s=x/3$
$sum_s(1/2)^s=frac{1-(1/2)^{s+1}}{1-1/2}$
answered Jul 19 '16 at 5:14
AndreiAndrei
13.2k21230
$endgroup$
add a comment |
$begingroup$
This particular curve can be also $f(x)=6(1-2^{-x/3})$. Here is how I guessed this function:
you go in steps of 3, so you might have something $x/3$.
At the first step the function increases by 3, then at the next step by 3/2, then 3/4. From here, I would guess that the formula is $$sum_s3left(frac{1}{2}right)^s,$$ where $s=x/3$
$sum_s(1/2)^s=frac{1-(1/2)^{s+1}}{1-1/2}$
answered Jul 19 '16 at 5:14
AndreiAndrei
13.2k21230
$endgroup$
add a comment |
$begingroup$
This particular curve can be also $f(x)=6(1-2^{-x/3})$. Here is how I guessed this function:
you go in steps of 3, so you might have something $x/3$.
At the first step the function increases by 3, then at the next step by 3/2, then 3/4. From here, I would guess that the formula is $$sum_s3left(frac{1}{2}right)^s,$$ where $s=x/3$
$sum_s(1/2)^s=frac{1-(1/2)^{s+1}}{1-1/2}$
answered Jul 19 '16 at 5:14
AndreiAndrei
13.2k21230
$endgroup$
This particular curve can be also $f(x)=6(1-2^{-x/3})$. Here is how I guessed this function:
you go in steps of 3, so you might have something $x/3$.
At the first step the function increases by 3, then at the next step by 3/2, then 3/4. From here, I would guess that the formula is $$sum_s3left(frac{1}{2}right)^s,$$ where $s=x/3$
$sum_s(1/2)^s=frac{1-(1/2)^{s+1}}{1-1/2}$
answered Jul 19 '16 at 5:14
AndreiAndrei
13.2k21230
answered Jul 19 '16 at 5:14
AndreiAndrei
13.2k21230
answered Jul 19 '16 at 5:14
AndreiAndrei
13.2k21230
answered Jul 19 '16 at 5:14
AndreiAndrei
13.2k21230
13.2k21230
add a comment |
add a comment |
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1
$begingroup$
What type of curve are you talking about ?
$endgroup$
– Vivek Kaushik
Jul 19 '16 at 3:29
1
$begingroup$
Sadly there are many curves that pass through these four points, so it requires more information such as "cubic polynomial" to make a unique determination.
$endgroup$
– hardmath
Jul 19 '16 at 3:39