Ant climbing on bush
$begingroup$
An ant is on the ground and trying to climb on a (straight) ivy bush 10m high. It crawls up 0.1m each night, but at day, the bush grows uniformly by 0.5m (in its entire height). Will the ant ever reach the top of the ivy? If yes, in how many days? If no, justify why.
Let's convert everything in centimeters.
I have made a table with the values of the bush height every night and every day, basis the rules.
The ant starts with 1000 cm height above and 0 cm below. Then it advances by 10 cm so it has 990 cm above and 10 cm below. Total height is (still) 1000 cm.
In the morning, the height of the bush is proportionally extended by 5% so the up value is now
$990*(5/1000)+990 = 1039.5$ and the down is $10*(5/1000)+10 = 10.5$
Total height is of course increased by 50 cm.
Then at night, first value is decreased by 10 and second is increased by 10. Height remains the same.
We continue this way and have the following values:
Bush up (cm) Bush down (cm) Total
Initially 1000 0 1000
1st night 990 10 1000
1st day 1039,5 10,5 1050
2nd night 1029,5 20,5 1050
2nd day 1078,52381 21,47619048 1100
3rd night 1068,52381 31,47619048 1100
3rd day 1117,093074 32,90692641 1150
4th night 1107,093074 42,90692641 1150
4th day 1155,227555 44,77244495 1200
5th night 1145,227555 54,77244495 1200
We continue this way and now we make a graph of the 3 columns, bush up, bush down and Total. We notice that the 3 curves are declining, which means that they will never intersect. This means that the ant will never reach the top. However, this is not correct. I was told that the problem has a positive reply. Where am I wrong?
Thank you very much in anticipation!
calculus
asked Jan 7 at 19:47
Khalil. JahromiKhalil. Jahromi
684
$endgroup$
add a comment |
$begingroup$
An ant is on the ground and trying to climb on a (straight) ivy bush 10m high. It crawls up 0.1m each night, but at day, the bush grows uniformly by 0.5m (in its entire height). Will the ant ever reach the top of the ivy? If yes, in how many days? If no, justify why.
Let's convert everything in centimeters.
I have made a table with the values of the bush height every night and every day, basis the rules.
The ant starts with 1000 cm height above and 0 cm below. Then it advances by 10 cm so it has 990 cm above and 10 cm below. Total height is (still) 1000 cm.
In the morning, the height of the bush is proportionally extended by 5% so the up value is now
$990*(5/1000)+990 = 1039.5$ and the down is $10*(5/1000)+10 = 10.5$
Total height is of course increased by 50 cm.
Then at night, first value is decreased by 10 and second is increased by 10. Height remains the same.
We continue this way and have the following values:
Bush up (cm) Bush down (cm) Total
Initially 1000 0 1000
1st night 990 10 1000
1st day 1039,5 10,5 1050
2nd night 1029,5 20,5 1050
2nd day 1078,52381 21,47619048 1100
3rd night 1068,52381 31,47619048 1100
3rd day 1117,093074 32,90692641 1150
4th night 1107,093074 42,90692641 1150
4th day 1155,227555 44,77244495 1200
5th night 1145,227555 54,77244495 1200
We continue this way and now we make a graph of the 3 columns, bush up, bush down and Total. We notice that the 3 curves are declining, which means that they will never intersect. This means that the ant will never reach the top. However, this is not correct. I was told that the problem has a positive reply. Where am I wrong?
Thank you very much in anticipation!
calculus
asked Jan 7 at 19:47
Khalil. JahromiKhalil. Jahromi
684
$endgroup$
$begingroup$
I feel like.. unless there's a fire and that ivy burns down to the ground, you are correct.
$endgroup$
– T. Fo
Jan 7 at 20:03
1
$begingroup$
Your calculations are correct. But as the ant climbs the percentage of the distance covered from bottom to top is declining every day. Eventually he will get to the point where there is sufficient bush behind him that he can start to close the distance, but it takes a long time for that to happen.
$endgroup$
– Doug M
Jan 7 at 20:03
$begingroup$
Well I continued my list and got ~2875 days. Can someone validate my result by using a formula? I did it in Excel so not sure if it's correct...
$endgroup$
– Khalil. Jahromi
Jan 7 at 20:33
add a comment |
$begingroup$
An ant is on the ground and trying to climb on a (straight) ivy bush 10m high. It crawls up 0.1m each night, but at day, the bush grows uniformly by 0.5m (in its entire height). Will the ant ever reach the top of the ivy? If yes, in how many days? If no, justify why.
Let's convert everything in centimeters.
I have made a table with the values of the bush height every night and every day, basis the rules.
The ant starts with 1000 cm height above and 0 cm below. Then it advances by 10 cm so it has 990 cm above and 10 cm below. Total height is (still) 1000 cm.
In the morning, the height of the bush is proportionally extended by 5% so the up value is now
$990*(5/1000)+990 = 1039.5$ and the down is $10*(5/1000)+10 = 10.5$
Total height is of course increased by 50 cm.
Then at night, first value is decreased by 10 and second is increased by 10. Height remains the same.
We continue this way and have the following values:
Bush up (cm) Bush down (cm) Total
Initially 1000 0 1000
1st night 990 10 1000
1st day 1039,5 10,5 1050
2nd night 1029,5 20,5 1050
2nd day 1078,52381 21,47619048 1100
3rd night 1068,52381 31,47619048 1100
3rd day 1117,093074 32,90692641 1150
4th night 1107,093074 42,90692641 1150
4th day 1155,227555 44,77244495 1200
5th night 1145,227555 54,77244495 1200
We continue this way and now we make a graph of the 3 columns, bush up, bush down and Total. We notice that the 3 curves are declining, which means that they will never intersect. This means that the ant will never reach the top. However, this is not correct. I was told that the problem has a positive reply. Where am I wrong?
Thank you very much in anticipation!
calculus
asked Jan 7 at 19:47
Khalil. JahromiKhalil. Jahromi
684
$endgroup$
An ant is on the ground and trying to climb on a (straight) ivy bush 10m high. It crawls up 0.1m each night, but at day, the bush grows uniformly by 0.5m (in its entire height). Will the ant ever reach the top of the ivy? If yes, in how many days? If no, justify why.
Let's convert everything in centimeters.
I have made a table with the values of the bush height every night and every day, basis the rules.
The ant starts with 1000 cm height above and 0 cm below. Then it advances by 10 cm so it has 990 cm above and 10 cm below. Total height is (still) 1000 cm.
In the morning, the height of the bush is proportionally extended by 5% so the up value is now
$990*(5/1000)+990 = 1039.5$ and the down is $10*(5/1000)+10 = 10.5$
Total height is of course increased by 50 cm.
Then at night, first value is decreased by 10 and second is increased by 10. Height remains the same.
We continue this way and have the following values:
Bush up (cm) Bush down (cm) Total
Initially 1000 0 1000
1st night 990 10 1000
1st day 1039,5 10,5 1050
2nd night 1029,5 20,5 1050
2nd day 1078,52381 21,47619048 1100
3rd night 1068,52381 31,47619048 1100
3rd day 1117,093074 32,90692641 1150
4th night 1107,093074 42,90692641 1150
4th day 1155,227555 44,77244495 1200
5th night 1145,227555 54,77244495 1200
We continue this way and now we make a graph of the 3 columns, bush up, bush down and Total. We notice that the 3 curves are declining, which means that they will never intersect. This means that the ant will never reach the top. However, this is not correct. I was told that the problem has a positive reply. Where am I wrong?
Thank you very much in anticipation!
calculus
calculus
asked Jan 7 at 19:47
Khalil. JahromiKhalil. Jahromi
684
asked Jan 7 at 19:47
Khalil. JahromiKhalil. Jahromi
684
asked Jan 7 at 19:47
Khalil. JahromiKhalil. Jahromi
684
asked Jan 7 at 19:47
Khalil. JahromiKhalil. Jahromi
684
asked Jan 7 at 19:47
Khalil. JahromiKhalil. Jahromi
684
684
$begingroup$
I feel like.. unless there's a fire and that ivy burns down to the ground, you are correct.
$endgroup$
– T. Fo
Jan 7 at 20:03
1
$begingroup$
Your calculations are correct. But as the ant climbs the percentage of the distance covered from bottom to top is declining every day. Eventually he will get to the point where there is sufficient bush behind him that he can start to close the distance, but it takes a long time for that to happen.
$endgroup$
– Doug M
Jan 7 at 20:03
$begingroup$
Well I continued my list and got ~2875 days. Can someone validate my result by using a formula? I did it in Excel so not sure if it's correct...
$endgroup$
– Khalil. Jahromi
Jan 7 at 20:33
add a comment |
$begingroup$
I feel like.. unless there's a fire and that ivy burns down to the ground, you are correct.
$endgroup$
– T. Fo
Jan 7 at 20:03
1
$begingroup$
Your calculations are correct. But as the ant climbs the percentage of the distance covered from bottom to top is declining every day. Eventually he will get to the point where there is sufficient bush behind him that he can start to close the distance, but it takes a long time for that to happen.
$endgroup$
– Doug M
Jan 7 at 20:03
$begingroup$
Well I continued my list and got ~2875 days. Can someone validate my result by using a formula? I did it in Excel so not sure if it's correct...
$endgroup$
– Khalil. Jahromi
Jan 7 at 20:33
$begingroup$
I feel like.. unless there's a fire and that ivy burns down to the ground, you are correct.
$endgroup$
– T. Fo
Jan 7 at 20:03
$begingroup$
I feel like.. unless there's a fire and that ivy burns down to the ground, you are correct.
$endgroup$
– T. Fo
Jan 7 at 20:03
1
1
$begingroup$
Your calculations are correct. But as the ant climbs the percentage of the distance covered from bottom to top is declining every day. Eventually he will get to the point where there is sufficient bush behind him that he can start to close the distance, but it takes a long time for that to happen.
$endgroup$
– Doug M
Jan 7 at 20:03
$begingroup$
Your calculations are correct. But as the ant climbs the percentage of the distance covered from bottom to top is declining every day. Eventually he will get to the point where there is sufficient bush behind him that he can start to close the distance, but it takes a long time for that to happen.
$endgroup$
– Doug M
Jan 7 at 20:03
$begingroup$
Well I continued my list and got ~2875 days. Can someone validate my result by using a formula? I did it in Excel so not sure if it's correct...
$endgroup$
– Khalil. Jahromi
Jan 7 at 20:33
$begingroup$
Well I continued my list and got ~2875 days. Can someone validate my result by using a formula? I did it in Excel so not sure if it's correct...
$endgroup$
– Khalil. Jahromi
Jan 7 at 20:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The total height of the bush after $d$ days is $h_d=(10 + 0.5d){text{m}}$. When the ant climbs $0.1{text{m}}$ at night, this constitutes a fraction $0.1 / h_d = frac{1}{100 + 5d}$ of the total height. (Of course, when the bush grows during the day, the ant's relative position on the bush is not affected.) So you just need to know for what value of $N$ the sum $sum_{d=1}^{N}frac{1}{100+5d}$ first exceeds $1$.
This is if the bush grows first, before the ant ever climbs... your original formulation gives the ant a head start relative to this, because you let the ant climb before the bush grows. In that case, you'd want to know when $sum_{d=1}^{N}frac{1}{100+5d}$ exceeds $0.99$. Using Excel, I find that the climb takes $2874$ days for the latter case (in agreement with OP) and $3023$ days for the former case.
answered Jan 8 at 14:18
mjqxxxxmjqxxxx
31.6k24186
$endgroup$
$begingroup$
Nice solution. (Looks like you want $i=d$ in your sum.)
$endgroup$
– mathmandan
Jan 8 at 16:31
add a comment |
$begingroup$
For simplicity, let $0.1m$ be the unit.
Assume that on night $n$ the ant is at height $a_n$ and the bush has height $b_n$ (all measured in units, $a_0=0$, $b_0=100$).
So the new height of the bush is $b_{n+1}=b_n+5$, which yields the arithmetic series $b_n=5n+100$.
The new position of the ant is $a_{n+1}= (a_n+1)cdot frac{b_n+5}{b_n}= (a_n+1)cdot frac{n+21}{n+20}$. (Easy calculation.)
As an auxiliary series, we introduce $x_n:= frac{a_n}{n+20}$. Then the above formula translates to $c_{n+1}= c_n+frac{1}{n+20}$.
Thus $c_n$ is almost the harmonic series.
(The harmonic series is $H_n= sumlimits_{k=1}^n frac{1}{k}$. It is well-known that $H_n>log n$.)
In fact $c_n= H_{n+20}-H_{20}$, and in particular $a_n=(H_{n+20}-H_{20})(n+20)$, which beats $b_n=5(n+20)$ as soon as $H_{n+20}-H_{20}geq 5$.
With precise calculation, this yields that the ant reaches the top exactly on day $3043$, so you must have made a rounding error with Excel.
answered Jan 7 at 21:53
A. PongráczA. Pongrácz
6,0821929
$endgroup$
$begingroup$
A. Pongracz, Thank you very much for your solution - but this leads to a different result than my simplistic method. Is there any way to solve this without harmonic series and logarithms, which is something I do not understand? This problem is supposed to be solved with elementary maths.
$endgroup$
– Khalil. Jahromi
Jan 8 at 9:32
$begingroup$
I think this is exactly what you asked for... You don't have to understand anything about harmonic series or logarithms. I gave you an EXPLICIT formula for $a_n$, namely $a_n= (H_{n+20}-H_{20})cdot (n+20)$. Here, $H_{20}$ is a concrete rational number. If you want to express $a_n$, you CANNOT avoid the harmonic series, as this formula is the correct, explicit formula, and it contains the harmonic series. By the way: what would be better than an explicit formula?
$endgroup$
– A. Pongrácz
Jan 8 at 13:08
$begingroup$
This explicit formula is particularly useful, because it shows that you have made a rounding error, and in fact 2875 is not a correct solution. I used wolframalpha: it showed that $a_{2875}leq 4.9435cdot (2875+20)$, which is strictly less than $5cdot (2875+20)= b_{2875}$. By using my explicit formula, you can easily find the best index. And you are much more protected against rounding errors. This answers to your question "Can someone validate my result by using a formula?". I believe that is exactly what I did (except the result was wrong.) So again: what kind of answer do you expect?
$endgroup$
– A. Pongrácz
Jan 8 at 13:12
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
The total height of the bush after $d$ days is $h_d=(10 + 0.5d){text{m}}$. When the ant climbs $0.1{text{m}}$ at night, this constitutes a fraction $0.1 / h_d = frac{1}{100 + 5d}$ of the total height. (Of course, when the bush grows during the day, the ant's relative position on the bush is not affected.) So you just need to know for what value of $N$ the sum $sum_{d=1}^{N}frac{1}{100+5d}$ first exceeds $1$.
This is if the bush grows first, before the ant ever climbs... your original formulation gives the ant a head start relative to this, because you let the ant climb before the bush grows. In that case, you'd want to know when $sum_{d=1}^{N}frac{1}{100+5d}$ exceeds $0.99$. Using Excel, I find that the climb takes $2874$ days for the latter case (in agreement with OP) and $3023$ days for the former case.
answered Jan 8 at 14:18
mjqxxxxmjqxxxx
31.6k24186
$endgroup$
$begingroup$
Nice solution. (Looks like you want $i=d$ in your sum.)
$endgroup$
– mathmandan
Jan 8 at 16:31
add a comment |
$begingroup$
The total height of the bush after $d$ days is $h_d=(10 + 0.5d){text{m}}$. When the ant climbs $0.1{text{m}}$ at night, this constitutes a fraction $0.1 / h_d = frac{1}{100 + 5d}$ of the total height. (Of course, when the bush grows during the day, the ant's relative position on the bush is not affected.) So you just need to know for what value of $N$ the sum $sum_{d=1}^{N}frac{1}{100+5d}$ first exceeds $1$.
This is if the bush grows first, before the ant ever climbs... your original formulation gives the ant a head start relative to this, because you let the ant climb before the bush grows. In that case, you'd want to know when $sum_{d=1}^{N}frac{1}{100+5d}$ exceeds $0.99$. Using Excel, I find that the climb takes $2874$ days for the latter case (in agreement with OP) and $3023$ days for the former case.
answered Jan 8 at 14:18
mjqxxxxmjqxxxx
31.6k24186
$endgroup$
$begingroup$
Nice solution. (Looks like you want $i=d$ in your sum.)
$endgroup$
– mathmandan
Jan 8 at 16:31
add a comment |
$begingroup$
The total height of the bush after $d$ days is $h_d=(10 + 0.5d){text{m}}$. When the ant climbs $0.1{text{m}}$ at night, this constitutes a fraction $0.1 / h_d = frac{1}{100 + 5d}$ of the total height. (Of course, when the bush grows during the day, the ant's relative position on the bush is not affected.) So you just need to know for what value of $N$ the sum $sum_{d=1}^{N}frac{1}{100+5d}$ first exceeds $1$.
This is if the bush grows first, before the ant ever climbs... your original formulation gives the ant a head start relative to this, because you let the ant climb before the bush grows. In that case, you'd want to know when $sum_{d=1}^{N}frac{1}{100+5d}$ exceeds $0.99$. Using Excel, I find that the climb takes $2874$ days for the latter case (in agreement with OP) and $3023$ days for the former case.
answered Jan 8 at 14:18
mjqxxxxmjqxxxx
31.6k24186
$endgroup$
The total height of the bush after $d$ days is $h_d=(10 + 0.5d){text{m}}$. When the ant climbs $0.1{text{m}}$ at night, this constitutes a fraction $0.1 / h_d = frac{1}{100 + 5d}$ of the total height. (Of course, when the bush grows during the day, the ant's relative position on the bush is not affected.) So you just need to know for what value of $N$ the sum $sum_{d=1}^{N}frac{1}{100+5d}$ first exceeds $1$.
This is if the bush grows first, before the ant ever climbs... your original formulation gives the ant a head start relative to this, because you let the ant climb before the bush grows. In that case, you'd want to know when $sum_{d=1}^{N}frac{1}{100+5d}$ exceeds $0.99$. Using Excel, I find that the climb takes $2874$ days for the latter case (in agreement with OP) and $3023$ days for the former case.
answered Jan 8 at 14:18
mjqxxxxmjqxxxx
31.6k24186
edited Jan 9 at 3:07
answered Jan 8 at 14:18
mjqxxxxmjqxxxx
31.6k24186
answered Jan 8 at 14:18
mjqxxxxmjqxxxx
31.6k24186
answered Jan 8 at 14:18
mjqxxxxmjqxxxx
31.6k24186
31.6k24186
$begingroup$
Nice solution. (Looks like you want $i=d$ in your sum.)
$endgroup$
– mathmandan
Jan 8 at 16:31
add a comment |
$begingroup$
Nice solution. (Looks like you want $i=d$ in your sum.)
$endgroup$
– mathmandan
Jan 8 at 16:31
$begingroup$
Nice solution. (Looks like you want $i=d$ in your sum.)
$endgroup$
– mathmandan
Jan 8 at 16:31
$begingroup$
Nice solution. (Looks like you want $i=d$ in your sum.)
$endgroup$
– mathmandan
Jan 8 at 16:31
add a comment |
$begingroup$
For simplicity, let $0.1m$ be the unit.
Assume that on night $n$ the ant is at height $a_n$ and the bush has height $b_n$ (all measured in units, $a_0=0$, $b_0=100$).
So the new height of the bush is $b_{n+1}=b_n+5$, which yields the arithmetic series $b_n=5n+100$.
The new position of the ant is $a_{n+1}= (a_n+1)cdot frac{b_n+5}{b_n}= (a_n+1)cdot frac{n+21}{n+20}$. (Easy calculation.)
As an auxiliary series, we introduce $x_n:= frac{a_n}{n+20}$. Then the above formula translates to $c_{n+1}= c_n+frac{1}{n+20}$.
Thus $c_n$ is almost the harmonic series.
(The harmonic series is $H_n= sumlimits_{k=1}^n frac{1}{k}$. It is well-known that $H_n>log n$.)
In fact $c_n= H_{n+20}-H_{20}$, and in particular $a_n=(H_{n+20}-H_{20})(n+20)$, which beats $b_n=5(n+20)$ as soon as $H_{n+20}-H_{20}geq 5$.
With precise calculation, this yields that the ant reaches the top exactly on day $3043$, so you must have made a rounding error with Excel.
answered Jan 7 at 21:53
A. PongráczA. Pongrácz
6,0821929
$endgroup$
$begingroup$
A. Pongracz, Thank you very much for your solution - but this leads to a different result than my simplistic method. Is there any way to solve this without harmonic series and logarithms, which is something I do not understand? This problem is supposed to be solved with elementary maths.
$endgroup$
– Khalil. Jahromi
Jan 8 at 9:32
$begingroup$
I think this is exactly what you asked for... You don't have to understand anything about harmonic series or logarithms. I gave you an EXPLICIT formula for $a_n$, namely $a_n= (H_{n+20}-H_{20})cdot (n+20)$. Here, $H_{20}$ is a concrete rational number. If you want to express $a_n$, you CANNOT avoid the harmonic series, as this formula is the correct, explicit formula, and it contains the harmonic series. By the way: what would be better than an explicit formula?
$endgroup$
– A. Pongrácz
Jan 8 at 13:08
$begingroup$
This explicit formula is particularly useful, because it shows that you have made a rounding error, and in fact 2875 is not a correct solution. I used wolframalpha: it showed that $a_{2875}leq 4.9435cdot (2875+20)$, which is strictly less than $5cdot (2875+20)= b_{2875}$. By using my explicit formula, you can easily find the best index. And you are much more protected against rounding errors. This answers to your question "Can someone validate my result by using a formula?". I believe that is exactly what I did (except the result was wrong.) So again: what kind of answer do you expect?
$endgroup$
– A. Pongrácz
Jan 8 at 13:12
add a comment |
$begingroup$
For simplicity, let $0.1m$ be the unit.
Assume that on night $n$ the ant is at height $a_n$ and the bush has height $b_n$ (all measured in units, $a_0=0$, $b_0=100$).
So the new height of the bush is $b_{n+1}=b_n+5$, which yields the arithmetic series $b_n=5n+100$.
The new position of the ant is $a_{n+1}= (a_n+1)cdot frac{b_n+5}{b_n}= (a_n+1)cdot frac{n+21}{n+20}$. (Easy calculation.)
As an auxiliary series, we introduce $x_n:= frac{a_n}{n+20}$. Then the above formula translates to $c_{n+1}= c_n+frac{1}{n+20}$.
Thus $c_n$ is almost the harmonic series.
(The harmonic series is $H_n= sumlimits_{k=1}^n frac{1}{k}$. It is well-known that $H_n>log n$.)
In fact $c_n= H_{n+20}-H_{20}$, and in particular $a_n=(H_{n+20}-H_{20})(n+20)$, which beats $b_n=5(n+20)$ as soon as $H_{n+20}-H_{20}geq 5$.
With precise calculation, this yields that the ant reaches the top exactly on day $3043$, so you must have made a rounding error with Excel.
answered Jan 7 at 21:53
A. PongráczA. Pongrácz
6,0821929
$endgroup$
$begingroup$
A. Pongracz, Thank you very much for your solution - but this leads to a different result than my simplistic method. Is there any way to solve this without harmonic series and logarithms, which is something I do not understand? This problem is supposed to be solved with elementary maths.
$endgroup$
– Khalil. Jahromi
Jan 8 at 9:32
$begingroup$
I think this is exactly what you asked for... You don't have to understand anything about harmonic series or logarithms. I gave you an EXPLICIT formula for $a_n$, namely $a_n= (H_{n+20}-H_{20})cdot (n+20)$. Here, $H_{20}$ is a concrete rational number. If you want to express $a_n$, you CANNOT avoid the harmonic series, as this formula is the correct, explicit formula, and it contains the harmonic series. By the way: what would be better than an explicit formula?
$endgroup$
– A. Pongrácz
Jan 8 at 13:08
$begingroup$
This explicit formula is particularly useful, because it shows that you have made a rounding error, and in fact 2875 is not a correct solution. I used wolframalpha: it showed that $a_{2875}leq 4.9435cdot (2875+20)$, which is strictly less than $5cdot (2875+20)= b_{2875}$. By using my explicit formula, you can easily find the best index. And you are much more protected against rounding errors. This answers to your question "Can someone validate my result by using a formula?". I believe that is exactly what I did (except the result was wrong.) So again: what kind of answer do you expect?
$endgroup$
– A. Pongrácz
Jan 8 at 13:12
add a comment |
$begingroup$
For simplicity, let $0.1m$ be the unit.
Assume that on night $n$ the ant is at height $a_n$ and the bush has height $b_n$ (all measured in units, $a_0=0$, $b_0=100$).
So the new height of the bush is $b_{n+1}=b_n+5$, which yields the arithmetic series $b_n=5n+100$.
The new position of the ant is $a_{n+1}= (a_n+1)cdot frac{b_n+5}{b_n}= (a_n+1)cdot frac{n+21}{n+20}$. (Easy calculation.)
As an auxiliary series, we introduce $x_n:= frac{a_n}{n+20}$. Then the above formula translates to $c_{n+1}= c_n+frac{1}{n+20}$.
Thus $c_n$ is almost the harmonic series.
(The harmonic series is $H_n= sumlimits_{k=1}^n frac{1}{k}$. It is well-known that $H_n>log n$.)
In fact $c_n= H_{n+20}-H_{20}$, and in particular $a_n=(H_{n+20}-H_{20})(n+20)$, which beats $b_n=5(n+20)$ as soon as $H_{n+20}-H_{20}geq 5$.
With precise calculation, this yields that the ant reaches the top exactly on day $3043$, so you must have made a rounding error with Excel.
answered Jan 7 at 21:53
A. PongráczA. Pongrácz
6,0821929
$endgroup$
For simplicity, let $0.1m$ be the unit.
Assume that on night $n$ the ant is at height $a_n$ and the bush has height $b_n$ (all measured in units, $a_0=0$, $b_0=100$).
So the new height of the bush is $b_{n+1}=b_n+5$, which yields the arithmetic series $b_n=5n+100$.
The new position of the ant is $a_{n+1}= (a_n+1)cdot frac{b_n+5}{b_n}= (a_n+1)cdot frac{n+21}{n+20}$. (Easy calculation.)
As an auxiliary series, we introduce $x_n:= frac{a_n}{n+20}$. Then the above formula translates to $c_{n+1}= c_n+frac{1}{n+20}$.
Thus $c_n$ is almost the harmonic series.
(The harmonic series is $H_n= sumlimits_{k=1}^n frac{1}{k}$. It is well-known that $H_n>log n$.)
In fact $c_n= H_{n+20}-H_{20}$, and in particular $a_n=(H_{n+20}-H_{20})(n+20)$, which beats $b_n=5(n+20)$ as soon as $H_{n+20}-H_{20}geq 5$.
With precise calculation, this yields that the ant reaches the top exactly on day $3043$, so you must have made a rounding error with Excel.
answered Jan 7 at 21:53
A. PongráczA. Pongrácz
6,0821929
edited Jan 8 at 13:21
answered Jan 7 at 21:53
A. PongráczA. Pongrácz
6,0821929
answered Jan 7 at 21:53
A. PongráczA. Pongrácz
6,0821929
answered Jan 7 at 21:53
A. PongráczA. Pongrácz
6,0821929
6,0821929
$begingroup$
A. Pongracz, Thank you very much for your solution - but this leads to a different result than my simplistic method. Is there any way to solve this without harmonic series and logarithms, which is something I do not understand? This problem is supposed to be solved with elementary maths.
$endgroup$
– Khalil. Jahromi
Jan 8 at 9:32
$begingroup$
I think this is exactly what you asked for... You don't have to understand anything about harmonic series or logarithms. I gave you an EXPLICIT formula for $a_n$, namely $a_n= (H_{n+20}-H_{20})cdot (n+20)$. Here, $H_{20}$ is a concrete rational number. If you want to express $a_n$, you CANNOT avoid the harmonic series, as this formula is the correct, explicit formula, and it contains the harmonic series. By the way: what would be better than an explicit formula?
$endgroup$
– A. Pongrácz
Jan 8 at 13:08
$begingroup$
This explicit formula is particularly useful, because it shows that you have made a rounding error, and in fact 2875 is not a correct solution. I used wolframalpha: it showed that $a_{2875}leq 4.9435cdot (2875+20)$, which is strictly less than $5cdot (2875+20)= b_{2875}$. By using my explicit formula, you can easily find the best index. And you are much more protected against rounding errors. This answers to your question "Can someone validate my result by using a formula?". I believe that is exactly what I did (except the result was wrong.) So again: what kind of answer do you expect?
$endgroup$
– A. Pongrácz
Jan 8 at 13:12
add a comment |
$begingroup$
A. Pongracz, Thank you very much for your solution - but this leads to a different result than my simplistic method. Is there any way to solve this without harmonic series and logarithms, which is something I do not understand? This problem is supposed to be solved with elementary maths.
$endgroup$
– Khalil. Jahromi
Jan 8 at 9:32
$begingroup$
I think this is exactly what you asked for... You don't have to understand anything about harmonic series or logarithms. I gave you an EXPLICIT formula for $a_n$, namely $a_n= (H_{n+20}-H_{20})cdot (n+20)$. Here, $H_{20}$ is a concrete rational number. If you want to express $a_n$, you CANNOT avoid the harmonic series, as this formula is the correct, explicit formula, and it contains the harmonic series. By the way: what would be better than an explicit formula?
$endgroup$
– A. Pongrácz
Jan 8 at 13:08
$begingroup$
This explicit formula is particularly useful, because it shows that you have made a rounding error, and in fact 2875 is not a correct solution. I used wolframalpha: it showed that $a_{2875}leq 4.9435cdot (2875+20)$, which is strictly less than $5cdot (2875+20)= b_{2875}$. By using my explicit formula, you can easily find the best index. And you are much more protected against rounding errors. This answers to your question "Can someone validate my result by using a formula?". I believe that is exactly what I did (except the result was wrong.) So again: what kind of answer do you expect?
$endgroup$
– A. Pongrácz
Jan 8 at 13:12
$begingroup$
A. Pongracz, Thank you very much for your solution - but this leads to a different result than my simplistic method. Is there any way to solve this without harmonic series and logarithms, which is something I do not understand? This problem is supposed to be solved with elementary maths.
$endgroup$
– Khalil. Jahromi
Jan 8 at 9:32
$begingroup$
A. Pongracz, Thank you very much for your solution - but this leads to a different result than my simplistic method. Is there any way to solve this without harmonic series and logarithms, which is something I do not understand? This problem is supposed to be solved with elementary maths.
$endgroup$
– Khalil. Jahromi
Jan 8 at 9:32
$begingroup$
I think this is exactly what you asked for... You don't have to understand anything about harmonic series or logarithms. I gave you an EXPLICIT formula for $a_n$, namely $a_n= (H_{n+20}-H_{20})cdot (n+20)$. Here, $H_{20}$ is a concrete rational number. If you want to express $a_n$, you CANNOT avoid the harmonic series, as this formula is the correct, explicit formula, and it contains the harmonic series. By the way: what would be better than an explicit formula?
$endgroup$
– A. Pongrácz
Jan 8 at 13:08
$begingroup$
I think this is exactly what you asked for... You don't have to understand anything about harmonic series or logarithms. I gave you an EXPLICIT formula for $a_n$, namely $a_n= (H_{n+20}-H_{20})cdot (n+20)$. Here, $H_{20}$ is a concrete rational number. If you want to express $a_n$, you CANNOT avoid the harmonic series, as this formula is the correct, explicit formula, and it contains the harmonic series. By the way: what would be better than an explicit formula?
$endgroup$
– A. Pongrácz
Jan 8 at 13:08
$begingroup$
This explicit formula is particularly useful, because it shows that you have made a rounding error, and in fact 2875 is not a correct solution. I used wolframalpha: it showed that $a_{2875}leq 4.9435cdot (2875+20)$, which is strictly less than $5cdot (2875+20)= b_{2875}$. By using my explicit formula, you can easily find the best index. And you are much more protected against rounding errors. This answers to your question "Can someone validate my result by using a formula?". I believe that is exactly what I did (except the result was wrong.) So again: what kind of answer do you expect?
$endgroup$
– A. Pongrácz
Jan 8 at 13:12
$begingroup$
This explicit formula is particularly useful, because it shows that you have made a rounding error, and in fact 2875 is not a correct solution. I used wolframalpha: it showed that $a_{2875}leq 4.9435cdot (2875+20)$, which is strictly less than $5cdot (2875+20)= b_{2875}$. By using my explicit formula, you can easily find the best index. And you are much more protected against rounding errors. This answers to your question "Can someone validate my result by using a formula?". I believe that is exactly what I did (except the result was wrong.) So again: what kind of answer do you expect?
$endgroup$
– A. Pongrácz
Jan 8 at 13:12
add a comment |
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$begingroup$
I feel like.. unless there's a fire and that ivy burns down to the ground, you are correct.
$endgroup$
– T. Fo
Jan 7 at 20:03
1
$begingroup$
Your calculations are correct. But as the ant climbs the percentage of the distance covered from bottom to top is declining every day. Eventually he will get to the point where there is sufficient bush behind him that he can start to close the distance, but it takes a long time for that to happen.
$endgroup$
– Doug M
Jan 7 at 20:03
$begingroup$
Well I continued my list and got ~2875 days. Can someone validate my result by using a formula? I did it in Excel so not sure if it's correct...
$endgroup$
– Khalil. Jahromi
Jan 7 at 20:33