Mixing
“Fill in the Gaps” Trig function with integer zeros another trig function doesn't have
I have an interesting challenge involving roots of trig functions. I'm wondering if there is a method of creating a function that hits the integer roots that $$sin(frac{pi}{5}x)*sin(frac{pi}{3}x)*sin(frac{pi}{4}x)$$ doesn't (i.e. instead of 3, 4, 5, 6, 8, 9, 10, 12..., it has roots at 2, 7, 11...). A generalized method for doing so is more of the thing I am asking for rather than the specific answer to this problem. Thanks for all help and creative answers!
functions trigonometry roots
edited Nov 22 '18 at 4:23
achille hui
95.6k5130257
asked Nov 22 '18 at 0:49
Ryan SheslerRyan Shesler
86
add a comment |
I have an interesting challenge involving roots of trig functions. I'm wondering if there is a method of creating a function that hits the integer roots that $$sin(frac{pi}{5}x)*sin(frac{pi}{3}x)*sin(frac{pi}{4}x)$$ doesn't (i.e. instead of 3, 4, 5, 6, 8, 9, 10, 12..., it has roots at 2, 7, 11...). A generalized method for doing so is more of the thing I am asking for rather than the specific answer to this problem. Thanks for all help and creative answers!
functions trigonometry roots
edited Nov 22 '18 at 4:23
achille hui
95.6k5130257
asked Nov 22 '18 at 0:49
Ryan SheslerRyan Shesler
86
2
rollback to previous version - vandalizing one's own question is a no-no on math.SE. If there is a real need, you can convince the mods to disassociate this question from you.
– achille hui
Nov 22 '18 at 4:29
add a comment |
I have an interesting challenge involving roots of trig functions. I'm wondering if there is a method of creating a function that hits the integer roots that $$sin(frac{pi}{5}x)*sin(frac{pi}{3}x)*sin(frac{pi}{4}x)$$ doesn't (i.e. instead of 3, 4, 5, 6, 8, 9, 10, 12..., it has roots at 2, 7, 11...). A generalized method for doing so is more of the thing I am asking for rather than the specific answer to this problem. Thanks for all help and creative answers!
functions trigonometry roots
edited Nov 22 '18 at 4:23
achille hui
95.6k5130257
asked Nov 22 '18 at 0:49
Ryan SheslerRyan Shesler
86
I have an interesting challenge involving roots of trig functions. I'm wondering if there is a method of creating a function that hits the integer roots that $$sin(frac{pi}{5}x)*sin(frac{pi}{3}x)*sin(frac{pi}{4}x)$$ doesn't (i.e. instead of 3, 4, 5, 6, 8, 9, 10, 12..., it has roots at 2, 7, 11...). A generalized method for doing so is more of the thing I am asking for rather than the specific answer to this problem. Thanks for all help and creative answers!
functions trigonometry roots
functions trigonometry roots
edited Nov 22 '18 at 4:23
achille hui
95.6k5130257
asked Nov 22 '18 at 0:49
Ryan SheslerRyan Shesler
86
edited Nov 22 '18 at 4:23
achille hui
95.6k5130257
asked Nov 22 '18 at 0:49
Ryan SheslerRyan Shesler
86
edited Nov 22 '18 at 4:23
achille hui
95.6k5130257
edited Nov 22 '18 at 4:23
achille hui
95.6k5130257
edited Nov 22 '18 at 4:23
achille hui
95.6k5130257
95.6k5130257
asked Nov 22 '18 at 0:49
Ryan SheslerRyan Shesler
86
asked Nov 22 '18 at 0:49
Ryan SheslerRyan Shesler
86
asked Nov 22 '18 at 0:49
Ryan SheslerRyan Shesler
86
86
2
rollback to previous version - vandalizing one's own question is a no-no on math.SE. If there is a real need, you can convince the mods to disassociate this question from you.
– achille hui
Nov 22 '18 at 4:29
add a comment |
2
rollback to previous version - vandalizing one's own question is a no-no on math.SE. If there is a real need, you can convince the mods to disassociate this question from you.
– achille hui
Nov 22 '18 at 4:29
2
2
rollback to previous version - vandalizing one's own question is a no-no on math.SE. If there is a real need, you can convince the mods to disassociate this question from you.
– achille hui
Nov 22 '18 at 4:29
rollback to previous version - vandalizing one's own question is a no-no on math.SE. If there is a real need, you can convince the mods to disassociate this question from you.
– achille hui
Nov 22 '18 at 4:29
add a comment |
1 Answer
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The function $$sinleft(frac{pi}{b}(x-a)right)$$ has roots $$..., a - b, a, a + b, a + 2b, ...$$
You can combine the roots of functions together by taking their product. So if you can break the roots you want into finitely many arithmetic sequences, you can a find pretty simple function that has those numbers as its roots. In your example, $$sinleft(frac{pi}{3} (x - 1)right) sinleft(frac{pi}{3} (x - 2)right)$$ fills in the gaps.
answered Nov 22 '18 at 0:59
WhatToDoWhatToDo
25116
1
Very nice. When formatting, use "sin" in mathjas to get $sin$ instead of $sin$.
– Ethan Bolker
Nov 22 '18 at 1:03
@EthanBolker Thank you.
– WhatToDo
Nov 22 '18 at 1:04
Thank you for the answer. The function I had in mind was a little more complicated, like $$sin(frac{pi}{2}x)*sin(frac{pi}{3}x)*sin(frac{pi}{4}x)$$
– Ryan Shesler
Nov 22 '18 at 2:07
@RyanShesler That function doesn't do what you said you wanted. It doesn't have a root at x=5, for example.
– WhatToDo
Nov 22 '18 at 2:10
Sorry I meant like this function has roots at 2, 3, 4, 6, 8, 9, 10. What's a way of finding another function with roots at 1, 5, 7, 11... Almost like a generalization of your answer to functions with more than one 'a'
– Ryan Shesler
Nov 22 '18 at 2:14
add a comment |
Your Answer
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1 Answer
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The function $$sinleft(frac{pi}{b}(x-a)right)$$ has roots $$..., a - b, a, a + b, a + 2b, ...$$
You can combine the roots of functions together by taking their product. So if you can break the roots you want into finitely many arithmetic sequences, you can a find pretty simple function that has those numbers as its roots. In your example, $$sinleft(frac{pi}{3} (x - 1)right) sinleft(frac{pi}{3} (x - 2)right)$$ fills in the gaps.
answered Nov 22 '18 at 0:59
WhatToDoWhatToDo
25116
1
Very nice. When formatting, use "sin" in mathjas to get $sin$ instead of $sin$.
– Ethan Bolker
Nov 22 '18 at 1:03
@EthanBolker Thank you.
– WhatToDo
Nov 22 '18 at 1:04
Thank you for the answer. The function I had in mind was a little more complicated, like $$sin(frac{pi}{2}x)*sin(frac{pi}{3}x)*sin(frac{pi}{4}x)$$
– Ryan Shesler
Nov 22 '18 at 2:07
@RyanShesler That function doesn't do what you said you wanted. It doesn't have a root at x=5, for example.
– WhatToDo
Nov 22 '18 at 2:10
Sorry I meant like this function has roots at 2, 3, 4, 6, 8, 9, 10. What's a way of finding another function with roots at 1, 5, 7, 11... Almost like a generalization of your answer to functions with more than one 'a'
– Ryan Shesler
Nov 22 '18 at 2:14
add a comment |
The function $$sinleft(frac{pi}{b}(x-a)right)$$ has roots $$..., a - b, a, a + b, a + 2b, ...$$
You can combine the roots of functions together by taking their product. So if you can break the roots you want into finitely many arithmetic sequences, you can a find pretty simple function that has those numbers as its roots. In your example, $$sinleft(frac{pi}{3} (x - 1)right) sinleft(frac{pi}{3} (x - 2)right)$$ fills in the gaps.
answered Nov 22 '18 at 0:59
WhatToDoWhatToDo
25116
1
Very nice. When formatting, use "sin" in mathjas to get $sin$ instead of $sin$.
– Ethan Bolker
Nov 22 '18 at 1:03
@EthanBolker Thank you.
– WhatToDo
Nov 22 '18 at 1:04
Thank you for the answer. The function I had in mind was a little more complicated, like $$sin(frac{pi}{2}x)*sin(frac{pi}{3}x)*sin(frac{pi}{4}x)$$
– Ryan Shesler
Nov 22 '18 at 2:07
@RyanShesler That function doesn't do what you said you wanted. It doesn't have a root at x=5, for example.
– WhatToDo
Nov 22 '18 at 2:10
Sorry I meant like this function has roots at 2, 3, 4, 6, 8, 9, 10. What's a way of finding another function with roots at 1, 5, 7, 11... Almost like a generalization of your answer to functions with more than one 'a'
– Ryan Shesler
Nov 22 '18 at 2:14
add a comment |
The function $$sinleft(frac{pi}{b}(x-a)right)$$ has roots $$..., a - b, a, a + b, a + 2b, ...$$
You can combine the roots of functions together by taking their product. So if you can break the roots you want into finitely many arithmetic sequences, you can a find pretty simple function that has those numbers as its roots. In your example, $$sinleft(frac{pi}{3} (x - 1)right) sinleft(frac{pi}{3} (x - 2)right)$$ fills in the gaps.
answered Nov 22 '18 at 0:59
WhatToDoWhatToDo
25116
The function $$sinleft(frac{pi}{b}(x-a)right)$$ has roots $$..., a - b, a, a + b, a + 2b, ...$$
You can combine the roots of functions together by taking their product. So if you can break the roots you want into finitely many arithmetic sequences, you can a find pretty simple function that has those numbers as its roots. In your example, $$sinleft(frac{pi}{3} (x - 1)right) sinleft(frac{pi}{3} (x - 2)right)$$ fills in the gaps.
answered Nov 22 '18 at 0:59
WhatToDoWhatToDo
25116
edited Nov 22 '18 at 1:18
answered Nov 22 '18 at 0:59
WhatToDoWhatToDo
25116
answered Nov 22 '18 at 0:59
WhatToDoWhatToDo
25116
answered Nov 22 '18 at 0:59
WhatToDoWhatToDo
25116
25116
1
Very nice. When formatting, use "sin" in mathjas to get $sin$ instead of $sin$.
– Ethan Bolker
Nov 22 '18 at 1:03
@EthanBolker Thank you.
– WhatToDo
Nov 22 '18 at 1:04
Thank you for the answer. The function I had in mind was a little more complicated, like $$sin(frac{pi}{2}x)*sin(frac{pi}{3}x)*sin(frac{pi}{4}x)$$
– Ryan Shesler
Nov 22 '18 at 2:07
@RyanShesler That function doesn't do what you said you wanted. It doesn't have a root at x=5, for example.
– WhatToDo
Nov 22 '18 at 2:10
Sorry I meant like this function has roots at 2, 3, 4, 6, 8, 9, 10. What's a way of finding another function with roots at 1, 5, 7, 11... Almost like a generalization of your answer to functions with more than one 'a'
– Ryan Shesler
Nov 22 '18 at 2:14
add a comment |
1
Very nice. When formatting, use "sin" in mathjas to get $sin$ instead of $sin$.
– Ethan Bolker
Nov 22 '18 at 1:03
@EthanBolker Thank you.
– WhatToDo
Nov 22 '18 at 1:04
Thank you for the answer. The function I had in mind was a little more complicated, like $$sin(frac{pi}{2}x)*sin(frac{pi}{3}x)*sin(frac{pi}{4}x)$$
– Ryan Shesler
Nov 22 '18 at 2:07
@RyanShesler That function doesn't do what you said you wanted. It doesn't have a root at x=5, for example.
– WhatToDo
Nov 22 '18 at 2:10
Sorry I meant like this function has roots at 2, 3, 4, 6, 8, 9, 10. What's a way of finding another function with roots at 1, 5, 7, 11... Almost like a generalization of your answer to functions with more than one 'a'
– Ryan Shesler
Nov 22 '18 at 2:14
1
1
Very nice. When formatting, use "sin" in mathjas to get $sin$ instead of $sin$.
– Ethan Bolker
Nov 22 '18 at 1:03
Very nice. When formatting, use "sin" in mathjas to get $sin$ instead of $sin$.
– Ethan Bolker
Nov 22 '18 at 1:03
@EthanBolker Thank you.
– WhatToDo
Nov 22 '18 at 1:04
@EthanBolker Thank you.
– WhatToDo
Nov 22 '18 at 1:04
Thank you for the answer. The function I had in mind was a little more complicated, like $$sin(frac{pi}{2}x)*sin(frac{pi}{3}x)*sin(frac{pi}{4}x)$$
– Ryan Shesler
Nov 22 '18 at 2:07
Thank you for the answer. The function I had in mind was a little more complicated, like $$sin(frac{pi}{2}x)*sin(frac{pi}{3}x)*sin(frac{pi}{4}x)$$
– Ryan Shesler
Nov 22 '18 at 2:07
@RyanShesler That function doesn't do what you said you wanted. It doesn't have a root at x=5, for example.
– WhatToDo
Nov 22 '18 at 2:10
@RyanShesler That function doesn't do what you said you wanted. It doesn't have a root at x=5, for example.
– WhatToDo
Nov 22 '18 at 2:10
Sorry I meant like this function has roots at 2, 3, 4, 6, 8, 9, 10. What's a way of finding another function with roots at 1, 5, 7, 11... Almost like a generalization of your answer to functions with more than one 'a'
– Ryan Shesler
Nov 22 '18 at 2:14
Sorry I meant like this function has roots at 2, 3, 4, 6, 8, 9, 10. What's a way of finding another function with roots at 1, 5, 7, 11... Almost like a generalization of your answer to functions with more than one 'a'
– Ryan Shesler
Nov 22 '18 at 2:14
add a comment |
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rollback to previous version - vandalizing one's own question is a no-no on math.SE. If there is a real need, you can convince the mods to disassociate this question from you.
– achille hui
Nov 22 '18 at 4:29