What is the maximum value of this fraction?
$begingroup$
If $x$ is positive, what is the maximum value of this expression:
$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$
This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.
This is what I have tried:
The denominator is a geometric series whose sum is
$$frac{1-x^{201}}{1-x}$$
The fraction now becomes
$$frac{x^{100}(1-x)}{1-x^{201}}$$
I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.
That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).
I can't see such means from looking at the fraction. Can someone help me here?
sequences-and-series algebra-precalculus means geometric-series
New contributor
$endgroup$
add a comment |
$begingroup$
If $x$ is positive, what is the maximum value of this expression:
$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$
This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.
This is what I have tried:
The denominator is a geometric series whose sum is
$$frac{1-x^{201}}{1-x}$$
The fraction now becomes
$$frac{x^{100}(1-x)}{1-x^{201}}$$
I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.
That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).
I can't see such means from looking at the fraction. Can someone help me here?
sequences-and-series algebra-precalculus means geometric-series
New contributor
$endgroup$
1
$begingroup$
For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
– Mark Viola
27 mins ago
add a comment |
$begingroup$
If $x$ is positive, what is the maximum value of this expression:
$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$
This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.
This is what I have tried:
The denominator is a geometric series whose sum is
$$frac{1-x^{201}}{1-x}$$
The fraction now becomes
$$frac{x^{100}(1-x)}{1-x^{201}}$$
I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.
That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).
I can't see such means from looking at the fraction. Can someone help me here?
sequences-and-series algebra-precalculus means geometric-series
New contributor
$endgroup$
If $x$ is positive, what is the maximum value of this expression:
$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$
This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.
This is what I have tried:
The denominator is a geometric series whose sum is
$$frac{1-x^{201}}{1-x}$$
The fraction now becomes
$$frac{x^{100}(1-x)}{1-x^{201}}$$
I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.
That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).
I can't see such means from looking at the fraction. Can someone help me here?
sequences-and-series algebra-precalculus means geometric-series
sequences-and-series algebra-precalculus means geometric-series
New contributor
New contributor
New contributor
asked 46 mins ago
user69284user69284
876
876
New contributor
New contributor
1
$begingroup$
For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
– Mark Viola
27 mins ago
add a comment |
1
$begingroup$
For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
– Mark Viola
27 mins ago
1
1
$begingroup$
For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
– Mark Viola
27 mins ago
$begingroup$
For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
– Mark Viola
27 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.
$endgroup$
add a comment |
$begingroup$
For $x>0$, we have from the AM-GM inequality
$$begin{align}
sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
&=201 sqrt[201]{x^{20100}}\\
&=201x^{100}
end{align}$$
Hence, we see that
$$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
10 mins ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.
$endgroup$
add a comment |
$begingroup$
The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.
$endgroup$
add a comment |
$begingroup$
The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.
$endgroup$
The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.
answered 30 mins ago
jmerryjmerry
5,957718
5,957718
add a comment |
add a comment |
$begingroup$
For $x>0$, we have from the AM-GM inequality
$$begin{align}
sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
&=201 sqrt[201]{x^{20100}}\\
&=201x^{100}
end{align}$$
Hence, we see that
$$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
10 mins ago
add a comment |
$begingroup$
For $x>0$, we have from the AM-GM inequality
$$begin{align}
sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
&=201 sqrt[201]{x^{20100}}\\
&=201x^{100}
end{align}$$
Hence, we see that
$$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
10 mins ago
add a comment |
$begingroup$
For $x>0$, we have from the AM-GM inequality
$$begin{align}
sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
&=201 sqrt[201]{x^{20100}}\\
&=201x^{100}
end{align}$$
Hence, we see that
$$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
For $x>0$, we have from the AM-GM inequality
$$begin{align}
sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
&=201 sqrt[201]{x^{20100}}\\
&=201x^{100}
end{align}$$
Hence, we see that
$$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
answered 24 mins ago
Mark ViolaMark Viola
131k1275171
131k1275171
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
10 mins ago
add a comment |
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
10 mins ago
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
10 mins ago
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
10 mins ago
add a comment |
user69284 is a new contributor. Be nice, and check out our Code of Conduct.
user69284 is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
– Mark Viola
27 mins ago