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Is there a continuous function satisfy following conditions?
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Is there a continuous function satisfy folloing conditions? and why?
$$int_0^1 f(x) dx = int_0^1 xf(x) dx = int_0^1 x^2f(x) dx = 0$$
with $f(0)=1$
functions
edited 11 mins ago
Tito Eliatron
359114
asked 58 mins ago
Dinzosm
83
add a comment |Â
up vote
1
down vote
favorite
Is there a continuous function satisfy folloing conditions? and why?
$$int_0^1 f(x) dx = int_0^1 xf(x) dx = int_0^1 x^2f(x) dx = 0$$
with $f(0)=1$
functions
edited 11 mins ago
Tito Eliatron
359114
asked 58 mins ago
Dinzosm
83
What are your thoughts? Is this an assignment give to you? Maybe there are some methods which were given to you recently which would work here?
– Yuriy S
49 mins ago
$f(x)=0$ is a trivial answer.
– Piquito
30 mins ago
@Piquito $f(0)$ must be 1.
– Tito Eliatron
29 mins ago
Right! Thanks you.
– Piquito
16 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there a continuous function satisfy folloing conditions? and why?
$$int_0^1 f(x) dx = int_0^1 xf(x) dx = int_0^1 x^2f(x) dx = 0$$
with $f(0)=1$
functions
edited 11 mins ago
Tito Eliatron
359114
asked 58 mins ago
Dinzosm
83
Is there a continuous function satisfy folloing conditions? and why?
$$int_0^1 f(x) dx = int_0^1 xf(x) dx = int_0^1 x^2f(x) dx = 0$$
with $f(0)=1$
functions
functions
edited 11 mins ago
Tito Eliatron
359114
asked 58 mins ago
Dinzosm
83
edited 11 mins ago
Tito Eliatron
359114
asked 58 mins ago
Dinzosm
83
edited 11 mins ago
Tito Eliatron
359114
edited 11 mins ago
Tito Eliatron
359114
edited 11 mins ago
Tito Eliatron
359114
359114
asked 58 mins ago
Dinzosm
83
asked 58 mins ago
Dinzosm
83
asked 58 mins ago
Dinzosm
83
83
What are your thoughts? Is this an assignment give to you? Maybe there are some methods which were given to you recently which would work here?
– Yuriy S
49 mins ago
$f(x)=0$ is a trivial answer.
– Piquito
30 mins ago
@Piquito $f(0)$ must be 1.
– Tito Eliatron
29 mins ago
Right! Thanks you.
– Piquito
16 mins ago
add a comment |Â
What are your thoughts? Is this an assignment give to you? Maybe there are some methods which were given to you recently which would work here?
– Yuriy S
49 mins ago
$f(x)=0$ is a trivial answer.
– Piquito
30 mins ago
@Piquito $f(0)$ must be 1.
– Tito Eliatron
29 mins ago
Right! Thanks you.
– Piquito
16 mins ago
What are your thoughts? Is this an assignment give to you? Maybe there are some methods which were given to you recently which would work here?
– Yuriy S
49 mins ago
What are your thoughts? Is this an assignment give to you? Maybe there are some methods which were given to you recently which would work here?
– Yuriy S
49 mins ago
$f(x)=0$ is a trivial answer.
– Piquito
30 mins ago
$f(x)=0$ is a trivial answer.
– Piquito
30 mins ago
@Piquito $f(0)$ must be 1.
– Tito Eliatron
29 mins ago
@Piquito $f(0)$ must be 1.
– Tito Eliatron
29 mins ago
Right! Thanks you.
– Piquito
16 mins ago
Right! Thanks you.
– Piquito
16 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
If you try $f(x)=ax^3+bx^2+cx+1$, then you have three linear equations in three unknowns.
$$beginalign
frac14a+frac13b+frac12c+1&=0\
frac15a+frac14b+frac13c+frac12&=0\
frac16a+frac15b+frac14c+frac13&=0
endalign$$
See if there is a solution to this system.
answered 32 mins ago
alex.jordan
37.6k559117
add a comment |Â
up vote
0
down vote
The most trivial function I can think of is $$fbegincases 1,x=0\0,xne0endcases$$
answered 37 mins ago
Andrei
8,9762923
I suppose $f$ is continuous he wants?
– AdditIdent
37 mins ago
Maybe, but that was not asked in the question
– Andrei
36 mins ago
I gave a similar suggestion to this earlier, but the OP seems to have updated their question.
– projectilemotion
18 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If you try $f(x)=ax^3+bx^2+cx+1$, then you have three linear equations in three unknowns.
$$beginalign
frac14a+frac13b+frac12c+1&=0\
frac15a+frac14b+frac13c+frac12&=0\
frac16a+frac15b+frac14c+frac13&=0
endalign$$
See if there is a solution to this system.
answered 32 mins ago
alex.jordan
37.6k559117
add a comment |Â
up vote
4
down vote
accepted
If you try $f(x)=ax^3+bx^2+cx+1$, then you have three linear equations in three unknowns.
$$beginalign
frac14a+frac13b+frac12c+1&=0\
frac15a+frac14b+frac13c+frac12&=0\
frac16a+frac15b+frac14c+frac13&=0
endalign$$
See if there is a solution to this system.
answered 32 mins ago
alex.jordan
37.6k559117
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If you try $f(x)=ax^3+bx^2+cx+1$, then you have three linear equations in three unknowns.
$$beginalign
frac14a+frac13b+frac12c+1&=0\
frac15a+frac14b+frac13c+frac12&=0\
frac16a+frac15b+frac14c+frac13&=0
endalign$$
See if there is a solution to this system.
answered 32 mins ago
alex.jordan
37.6k559117
If you try $f(x)=ax^3+bx^2+cx+1$, then you have three linear equations in three unknowns.
$$beginalign
frac14a+frac13b+frac12c+1&=0\
frac15a+frac14b+frac13c+frac12&=0\
frac16a+frac15b+frac14c+frac13&=0
endalign$$
See if there is a solution to this system.
answered 32 mins ago
alex.jordan
37.6k559117
answered 32 mins ago
alex.jordan
37.6k559117
answered 32 mins ago
alex.jordan
37.6k559117
answered 32 mins ago
alex.jordan
37.6k559117
37.6k559117
add a comment |Â
add a comment |Â
up vote
0
down vote
The most trivial function I can think of is $$fbegincases 1,x=0\0,xne0endcases$$
answered 37 mins ago
Andrei
8,9762923
I suppose $f$ is continuous he wants?
– AdditIdent
37 mins ago
Maybe, but that was not asked in the question
– Andrei
36 mins ago
I gave a similar suggestion to this earlier, but the OP seems to have updated their question.
– projectilemotion
18 mins ago
add a comment |Â
up vote
0
down vote
The most trivial function I can think of is $$fbegincases 1,x=0\0,xne0endcases$$
answered 37 mins ago
Andrei
8,9762923
I suppose $f$ is continuous he wants?
– AdditIdent
37 mins ago
Maybe, but that was not asked in the question
– Andrei
36 mins ago
I gave a similar suggestion to this earlier, but the OP seems to have updated their question.
– projectilemotion
18 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The most trivial function I can think of is $$fbegincases 1,x=0\0,xne0endcases$$
answered 37 mins ago
Andrei
8,9762923
The most trivial function I can think of is $$fbegincases 1,x=0\0,xne0endcases$$
answered 37 mins ago
Andrei
8,9762923
answered 37 mins ago
Andrei
8,9762923
answered 37 mins ago
Andrei
8,9762923
answered 37 mins ago
Andrei
8,9762923
8,9762923
I suppose $f$ is continuous he wants?
– AdditIdent
37 mins ago
Maybe, but that was not asked in the question
– Andrei
36 mins ago
I gave a similar suggestion to this earlier, but the OP seems to have updated their question.
– projectilemotion
18 mins ago
add a comment |Â
I suppose $f$ is continuous he wants?
– AdditIdent
37 mins ago
Maybe, but that was not asked in the question
– Andrei
36 mins ago
I gave a similar suggestion to this earlier, but the OP seems to have updated their question.
– projectilemotion
18 mins ago
I suppose $f$ is continuous he wants?
– AdditIdent
37 mins ago
I suppose $f$ is continuous he wants?
– AdditIdent
37 mins ago
Maybe, but that was not asked in the question
– Andrei
36 mins ago
Maybe, but that was not asked in the question
– Andrei
36 mins ago
I gave a similar suggestion to this earlier, but the OP seems to have updated their question.
– projectilemotion
18 mins ago
I gave a similar suggestion to this earlier, but the OP seems to have updated their question.
– projectilemotion
18 mins ago
add a comment |Â
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What are your thoughts? Is this an assignment give to you? Maybe there are some methods which were given to you recently which would work here?
– Yuriy S
49 mins ago
$f(x)=0$ is a trivial answer.
– Piquito
30 mins ago
@Piquito $f(0)$ must be 1.
– Tito Eliatron
29 mins ago
Right! Thanks you.
– Piquito
16 mins ago