Confused by notation for Mathematical Induction question
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The question says to allow $x_1=3$ and given $x_n$:
$$x_n+1 = frac14x_n + 9.$$Use mathematical induction to prove that for all $n inmathbb N$:
(a) $x_n+1> x_n$,
(b) $x_n < 12$.
I am not completely sure I know what's being asked here. Could someone clarify and give a hint.. I think I at least got the base case.. Compare the equation with 3 and then the equation with 4. Or is that wrong, too?
proof-verification induction
edited Sep 10 at 19:52
EuklidAlexandria
2979
asked Sep 10 at 19:23
NewbsMcgee
11
 |Â
show 2 more comments
up vote
0
down vote
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The question says to allow $x_1=3$ and given $x_n$:
$$x_n+1 = frac14x_n + 9.$$Use mathematical induction to prove that for all $n inmathbb N$:
(a) $x_n+1> x_n$,
(b) $x_n < 12$.
I am not completely sure I know what's being asked here. Could someone clarify and give a hint.. I think I at least got the base case.. Compare the equation with 3 and then the equation with 4. Or is that wrong, too?
proof-verification induction
edited Sep 10 at 19:52
EuklidAlexandria
2979
asked Sep 10 at 19:23
NewbsMcgee
11
You are asked to use mathematical induction. Have a look on this site (google mathematical induction, and the formula). For $n=1$, did you verify both statements already?
– Dietrich Burde
Sep 10 at 19:25
1
Is "1/4x<sub>n</sub>" supposed to be $frac14 x_n$ or $frac14x_n$?
– md2perpe
Sep 10 at 19:27
The former rather than the latter
– NewbsMcgee
Sep 10 at 19:59
for the base case, if $x_1=3$ can you get $x_2$ from the equation? What is $x_2$ in terms of $x_1$?
– Daniel Gendin
Sep 10 at 20:07
To save yourself some work note that $x_n+1 - x_n = frac34left(12 - x_nright)$ so proving $x_n+1 > x_n$ is the same as proving $12 > x_n$.
– Winther
Sep 10 at 20:18
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question says to allow $x_1=3$ and given $x_n$:
$$x_n+1 = frac14x_n + 9.$$Use mathematical induction to prove that for all $n inmathbb N$:
(a) $x_n+1> x_n$,
(b) $x_n < 12$.
I am not completely sure I know what's being asked here. Could someone clarify and give a hint.. I think I at least got the base case.. Compare the equation with 3 and then the equation with 4. Or is that wrong, too?
proof-verification induction
edited Sep 10 at 19:52
EuklidAlexandria
2979
asked Sep 10 at 19:23
NewbsMcgee
11
The question says to allow $x_1=3$ and given $x_n$:
$$x_n+1 = frac14x_n + 9.$$Use mathematical induction to prove that for all $n inmathbb N$:
(a) $x_n+1> x_n$,
(b) $x_n < 12$.
I am not completely sure I know what's being asked here. Could someone clarify and give a hint.. I think I at least got the base case.. Compare the equation with 3 and then the equation with 4. Or is that wrong, too?
proof-verification induction
proof-verification induction
edited Sep 10 at 19:52
EuklidAlexandria
2979
asked Sep 10 at 19:23
NewbsMcgee
11
edited Sep 10 at 19:52
EuklidAlexandria
2979
asked Sep 10 at 19:23
NewbsMcgee
11
edited Sep 10 at 19:52
EuklidAlexandria
2979
edited Sep 10 at 19:52
EuklidAlexandria
2979
edited Sep 10 at 19:52
EuklidAlexandria
2979
2979
asked Sep 10 at 19:23
NewbsMcgee
11
asked Sep 10 at 19:23
NewbsMcgee
11
asked Sep 10 at 19:23
NewbsMcgee
11
11
You are asked to use mathematical induction. Have a look on this site (google mathematical induction, and the formula). For $n=1$, did you verify both statements already?
– Dietrich Burde
Sep 10 at 19:25
1
Is "1/4x<sub>n</sub>" supposed to be $frac14 x_n$ or $frac14x_n$?
– md2perpe
Sep 10 at 19:27
The former rather than the latter
– NewbsMcgee
Sep 10 at 19:59
for the base case, if $x_1=3$ can you get $x_2$ from the equation? What is $x_2$ in terms of $x_1$?
– Daniel Gendin
Sep 10 at 20:07
To save yourself some work note that $x_n+1 - x_n = frac34left(12 - x_nright)$ so proving $x_n+1 > x_n$ is the same as proving $12 > x_n$.
– Winther
Sep 10 at 20:18
 |Â
show 2 more comments
You are asked to use mathematical induction. Have a look on this site (google mathematical induction, and the formula). For $n=1$, did you verify both statements already?
– Dietrich Burde
Sep 10 at 19:25
1
Is "1/4x<sub>n</sub>" supposed to be $frac14 x_n$ or $frac14x_n$?
– md2perpe
Sep 10 at 19:27
The former rather than the latter
– NewbsMcgee
Sep 10 at 19:59
for the base case, if $x_1=3$ can you get $x_2$ from the equation? What is $x_2$ in terms of $x_1$?
– Daniel Gendin
Sep 10 at 20:07
To save yourself some work note that $x_n+1 - x_n = frac34left(12 - x_nright)$ so proving $x_n+1 > x_n$ is the same as proving $12 > x_n$.
– Winther
Sep 10 at 20:18
You are asked to use mathematical induction. Have a look on this site (google mathematical induction, and the formula). For $n=1$, did you verify both statements already?
– Dietrich Burde
Sep 10 at 19:25
You are asked to use mathematical induction. Have a look on this site (google mathematical induction, and the formula). For $n=1$, did you verify both statements already?
– Dietrich Burde
Sep 10 at 19:25
1
1
Is "1/4x<sub>n</sub>" supposed to be $frac14 x_n$ or $frac14x_n$?
– md2perpe
Sep 10 at 19:27
Is "1/4x<sub>n</sub>" supposed to be $frac14 x_n$ or $frac14x_n$?
– md2perpe
Sep 10 at 19:27
The former rather than the latter
– NewbsMcgee
Sep 10 at 19:59
The former rather than the latter
– NewbsMcgee
Sep 10 at 19:59
for the base case, if $x_1=3$ can you get $x_2$ from the equation? What is $x_2$ in terms of $x_1$?
– Daniel Gendin
Sep 10 at 20:07
for the base case, if $x_1=3$ can you get $x_2$ from the equation? What is $x_2$ in terms of $x_1$?
– Daniel Gendin
Sep 10 at 20:07
To save yourself some work note that $x_n+1 - x_n = frac34left(12 - x_nright)$ so proving $x_n+1 > x_n$ is the same as proving $12 > x_n$.
– Winther
Sep 10 at 20:18
To save yourself some work note that $x_n+1 - x_n = frac34left(12 - x_nright)$ so proving $x_n+1 > x_n$ is the same as proving $12 > x_n$.
– Winther
Sep 10 at 20:18
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
I would start first by proving part b. If $x_n-1<12$ then $frac14x_n-1<3$, therefore $x_n<3+9=12$.
For part a, just compute $x_n+1-x_n$. You will get $9-frac34x_n$. But now you know $x_n<12$, so $x_n+1-x_n>9-frac3412$
answered Sep 10 at 20:18
Andrei
7,9852923
Hm.. I like the idea of that approach, but can we back up for a minute. For the base case, I was under the impression that I would just compare $x_1$ (3) and $x_2$ which is $1/4*3+9$ and that would be sufficient to prove the base case. Is that correct? And how did $x_n+1 - x_n = 9-(3/4)x_n$ Edited because I am trying to be better about formatting in latex or whatever it's called
– NewbsMcgee
Sep 11 at 1:02
Yes. I skipped that base case step. The general case is tricky. I do not know how to prove a without b.
– Andrei
Sep 11 at 3:10
Yeah, I absolutely think you're correct about b and I emailed my professor. He basically said as much. Could you just elaborate a little more on why $x_n < 3 + 9$? I understand that $x_n-1 < 12$, but I don't understand why we divided by 4 and where exactly the 9 came from (unless we only did that because we found that $1/4 x_n-1 < 3$ in which case that makes sense).
– NewbsMcgee
Sep 11 at 3:15
That's exactly why.
– Andrei
Sep 11 at 3:28
Thanks! I really appreciate the help. I've done mathematical induction before, but it's usually on sums and series. This is very different than what I have seen before.
– NewbsMcgee
Sep 11 at 3:32
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I would start first by proving part b. If $x_n-1<12$ then $frac14x_n-1<3$, therefore $x_n<3+9=12$.
For part a, just compute $x_n+1-x_n$. You will get $9-frac34x_n$. But now you know $x_n<12$, so $x_n+1-x_n>9-frac3412$
answered Sep 10 at 20:18
Andrei
7,9852923
Hm.. I like the idea of that approach, but can we back up for a minute. For the base case, I was under the impression that I would just compare $x_1$ (3) and $x_2$ which is $1/4*3+9$ and that would be sufficient to prove the base case. Is that correct? And how did $x_n+1 - x_n = 9-(3/4)x_n$ Edited because I am trying to be better about formatting in latex or whatever it's called
– NewbsMcgee
Sep 11 at 1:02
Yes. I skipped that base case step. The general case is tricky. I do not know how to prove a without b.
– Andrei
Sep 11 at 3:10
Yeah, I absolutely think you're correct about b and I emailed my professor. He basically said as much. Could you just elaborate a little more on why $x_n < 3 + 9$? I understand that $x_n-1 < 12$, but I don't understand why we divided by 4 and where exactly the 9 came from (unless we only did that because we found that $1/4 x_n-1 < 3$ in which case that makes sense).
– NewbsMcgee
Sep 11 at 3:15
That's exactly why.
– Andrei
Sep 11 at 3:28
Thanks! I really appreciate the help. I've done mathematical induction before, but it's usually on sums and series. This is very different than what I have seen before.
– NewbsMcgee
Sep 11 at 3:32
 |Â
show 2 more comments
up vote
0
down vote
accepted
I would start first by proving part b. If $x_n-1<12$ then $frac14x_n-1<3$, therefore $x_n<3+9=12$.
For part a, just compute $x_n+1-x_n$. You will get $9-frac34x_n$. But now you know $x_n<12$, so $x_n+1-x_n>9-frac3412$
answered Sep 10 at 20:18
Andrei
7,9852923
Hm.. I like the idea of that approach, but can we back up for a minute. For the base case, I was under the impression that I would just compare $x_1$ (3) and $x_2$ which is $1/4*3+9$ and that would be sufficient to prove the base case. Is that correct? And how did $x_n+1 - x_n = 9-(3/4)x_n$ Edited because I am trying to be better about formatting in latex or whatever it's called
– NewbsMcgee
Sep 11 at 1:02
Yes. I skipped that base case step. The general case is tricky. I do not know how to prove a without b.
– Andrei
Sep 11 at 3:10
Yeah, I absolutely think you're correct about b and I emailed my professor. He basically said as much. Could you just elaborate a little more on why $x_n < 3 + 9$? I understand that $x_n-1 < 12$, but I don't understand why we divided by 4 and where exactly the 9 came from (unless we only did that because we found that $1/4 x_n-1 < 3$ in which case that makes sense).
– NewbsMcgee
Sep 11 at 3:15
That's exactly why.
– Andrei
Sep 11 at 3:28
Thanks! I really appreciate the help. I've done mathematical induction before, but it's usually on sums and series. This is very different than what I have seen before.
– NewbsMcgee
Sep 11 at 3:32
 |Â
show 2 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I would start first by proving part b. If $x_n-1<12$ then $frac14x_n-1<3$, therefore $x_n<3+9=12$.
For part a, just compute $x_n+1-x_n$. You will get $9-frac34x_n$. But now you know $x_n<12$, so $x_n+1-x_n>9-frac3412$
answered Sep 10 at 20:18
Andrei
7,9852923
I would start first by proving part b. If $x_n-1<12$ then $frac14x_n-1<3$, therefore $x_n<3+9=12$.
For part a, just compute $x_n+1-x_n$. You will get $9-frac34x_n$. But now you know $x_n<12$, so $x_n+1-x_n>9-frac3412$
answered Sep 10 at 20:18
Andrei
7,9852923
answered Sep 10 at 20:18
Andrei
7,9852923
answered Sep 10 at 20:18
Andrei
7,9852923
answered Sep 10 at 20:18
Andrei
7,9852923
7,9852923
Hm.. I like the idea of that approach, but can we back up for a minute. For the base case, I was under the impression that I would just compare $x_1$ (3) and $x_2$ which is $1/4*3+9$ and that would be sufficient to prove the base case. Is that correct? And how did $x_n+1 - x_n = 9-(3/4)x_n$ Edited because I am trying to be better about formatting in latex or whatever it's called
– NewbsMcgee
Sep 11 at 1:02
Yes. I skipped that base case step. The general case is tricky. I do not know how to prove a without b.
– Andrei
Sep 11 at 3:10
Yeah, I absolutely think you're correct about b and I emailed my professor. He basically said as much. Could you just elaborate a little more on why $x_n < 3 + 9$? I understand that $x_n-1 < 12$, but I don't understand why we divided by 4 and where exactly the 9 came from (unless we only did that because we found that $1/4 x_n-1 < 3$ in which case that makes sense).
– NewbsMcgee
Sep 11 at 3:15
That's exactly why.
– Andrei
Sep 11 at 3:28
Thanks! I really appreciate the help. I've done mathematical induction before, but it's usually on sums and series. This is very different than what I have seen before.
– NewbsMcgee
Sep 11 at 3:32
 |Â
show 2 more comments
Hm.. I like the idea of that approach, but can we back up for a minute. For the base case, I was under the impression that I would just compare $x_1$ (3) and $x_2$ which is $1/4*3+9$ and that would be sufficient to prove the base case. Is that correct? And how did $x_n+1 - x_n = 9-(3/4)x_n$ Edited because I am trying to be better about formatting in latex or whatever it's called
– NewbsMcgee
Sep 11 at 1:02
Yes. I skipped that base case step. The general case is tricky. I do not know how to prove a without b.
– Andrei
Sep 11 at 3:10
Yeah, I absolutely think you're correct about b and I emailed my professor. He basically said as much. Could you just elaborate a little more on why $x_n < 3 + 9$? I understand that $x_n-1 < 12$, but I don't understand why we divided by 4 and where exactly the 9 came from (unless we only did that because we found that $1/4 x_n-1 < 3$ in which case that makes sense).
– NewbsMcgee
Sep 11 at 3:15
That's exactly why.
– Andrei
Sep 11 at 3:28
Thanks! I really appreciate the help. I've done mathematical induction before, but it's usually on sums and series. This is very different than what I have seen before.
– NewbsMcgee
Sep 11 at 3:32
Hm.. I like the idea of that approach, but can we back up for a minute. For the base case, I was under the impression that I would just compare $x_1$ (3) and $x_2$ which is $1/4*3+9$ and that would be sufficient to prove the base case. Is that correct? And how did $x_n+1 - x_n = 9-(3/4)x_n$ Edited because I am trying to be better about formatting in latex or whatever it's called
– NewbsMcgee
Sep 11 at 1:02
Hm.. I like the idea of that approach, but can we back up for a minute. For the base case, I was under the impression that I would just compare $x_1$ (3) and $x_2$ which is $1/4*3+9$ and that would be sufficient to prove the base case. Is that correct? And how did $x_n+1 - x_n = 9-(3/4)x_n$ Edited because I am trying to be better about formatting in latex or whatever it's called
– NewbsMcgee
Sep 11 at 1:02
Yes. I skipped that base case step. The general case is tricky. I do not know how to prove a without b.
– Andrei
Sep 11 at 3:10
Yes. I skipped that base case step. The general case is tricky. I do not know how to prove a without b.
– Andrei
Sep 11 at 3:10
Yeah, I absolutely think you're correct about b and I emailed my professor. He basically said as much. Could you just elaborate a little more on why $x_n < 3 + 9$? I understand that $x_n-1 < 12$, but I don't understand why we divided by 4 and where exactly the 9 came from (unless we only did that because we found that $1/4 x_n-1 < 3$ in which case that makes sense).
– NewbsMcgee
Sep 11 at 3:15
Yeah, I absolutely think you're correct about b and I emailed my professor. He basically said as much. Could you just elaborate a little more on why $x_n < 3 + 9$? I understand that $x_n-1 < 12$, but I don't understand why we divided by 4 and where exactly the 9 came from (unless we only did that because we found that $1/4 x_n-1 < 3$ in which case that makes sense).
– NewbsMcgee
Sep 11 at 3:15
That's exactly why.
– Andrei
Sep 11 at 3:28
That's exactly why.
– Andrei
Sep 11 at 3:28
Thanks! I really appreciate the help. I've done mathematical induction before, but it's usually on sums and series. This is very different than what I have seen before.
– NewbsMcgee
Sep 11 at 3:32
Thanks! I really appreciate the help. I've done mathematical induction before, but it's usually on sums and series. This is very different than what I have seen before.
– NewbsMcgee
Sep 11 at 3:32
 |Â
show 2 more comments
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You are asked to use mathematical induction. Have a look on this site (google mathematical induction, and the formula). For $n=1$, did you verify both statements already?
– Dietrich Burde
Sep 10 at 19:25
1
Is "1/4x<sub>n</sub>" supposed to be $frac14 x_n$ or $frac14x_n$?
– md2perpe
Sep 10 at 19:27
The former rather than the latter
– NewbsMcgee
Sep 10 at 19:59
for the base case, if $x_1=3$ can you get $x_2$ from the equation? What is $x_2$ in terms of $x_1$?
– Daniel Gendin
Sep 10 at 20:07
To save yourself some work note that $x_n+1 - x_n = frac34left(12 - x_nright)$ so proving $x_n+1 > x_n$ is the same as proving $12 > x_n$.
– Winther
Sep 10 at 20:18