Mixing
Prove a recursive function $f(p_i) = (p_i/2)g(p_i) to infty$ as $p_i to infty$?
$begingroup$
Let's say we have a set of numbers ${5,9,13,17,21,ldots, 5+4i,ldots}$ and each $p_i$ is a member of this set, namely $p_0 = 5, p_1 = 9, p_2 = 13$ , etc.
Consider the following functions defined recursively:
begin{align}
g(p_i) &= frac{p_i - 2}{p_i} g(p_{i-1}) & g(5) &= frac35 \
f(p_i) &= frac{ p_i}2 g(p_i) &&
end{align}
How does one prove that $f(p_i)$ goes to infinity as $p_i$ goes to infinity?
sequences-and-series limits discrete-mathematics recursion computational-mathematics
edited Feb 2 at 2:54
Lee David Chung Lin
4,47841242
asked Feb 1 at 21:57
temp wattstemp watts
1126
$endgroup$
add a comment |
$begingroup$
Let's say we have a set of numbers ${5,9,13,17,21,ldots, 5+4i,ldots}$ and each $p_i$ is a member of this set, namely $p_0 = 5, p_1 = 9, p_2 = 13$ , etc.
Consider the following functions defined recursively:
begin{align}
g(p_i) &= frac{p_i - 2}{p_i} g(p_{i-1}) & g(5) &= frac35 \
f(p_i) &= frac{ p_i}2 g(p_i) &&
end{align}
How does one prove that $f(p_i)$ goes to infinity as $p_i$ goes to infinity?
sequences-and-series limits discrete-mathematics recursion computational-mathematics
edited Feb 2 at 2:54
Lee David Chung Lin
4,47841242
asked Feb 1 at 21:57
temp wattstemp watts
1126
$endgroup$
$begingroup$
I would try computing the first values, and then try to prove, using induction on $i$ a statement of the form "for all $igeq 5$, $f(p_i)geq ci$ for some constant $c$ (that you would somehow guess computing the first terms).
$endgroup$
– J.F
Feb 1 at 22:04
add a comment |
$begingroup$
Let's say we have a set of numbers ${5,9,13,17,21,ldots, 5+4i,ldots}$ and each $p_i$ is a member of this set, namely $p_0 = 5, p_1 = 9, p_2 = 13$ , etc.
Consider the following functions defined recursively:
begin{align}
g(p_i) &= frac{p_i - 2}{p_i} g(p_{i-1}) & g(5) &= frac35 \
f(p_i) &= frac{ p_i}2 g(p_i) &&
end{align}
How does one prove that $f(p_i)$ goes to infinity as $p_i$ goes to infinity?
sequences-and-series limits discrete-mathematics recursion computational-mathematics
edited Feb 2 at 2:54
Lee David Chung Lin
4,47841242
asked Feb 1 at 21:57
temp wattstemp watts
1126
$endgroup$
Let's say we have a set of numbers ${5,9,13,17,21,ldots, 5+4i,ldots}$ and each $p_i$ is a member of this set, namely $p_0 = 5, p_1 = 9, p_2 = 13$ , etc.
Consider the following functions defined recursively:
begin{align}
g(p_i) &= frac{p_i - 2}{p_i} g(p_{i-1}) & g(5) &= frac35 \
f(p_i) &= frac{ p_i}2 g(p_i) &&
end{align}
How does one prove that $f(p_i)$ goes to infinity as $p_i$ goes to infinity?
sequences-and-series limits discrete-mathematics recursion computational-mathematics
sequences-and-series limits discrete-mathematics recursion computational-mathematics
edited Feb 2 at 2:54
Lee David Chung Lin
4,47841242
asked Feb 1 at 21:57
temp wattstemp watts
1126
edited Feb 2 at 2:54
Lee David Chung Lin
4,47841242
asked Feb 1 at 21:57
temp wattstemp watts
1126
edited Feb 2 at 2:54
Lee David Chung Lin
4,47841242
edited Feb 2 at 2:54
Lee David Chung Lin
4,47841242
edited Feb 2 at 2:54
Lee David Chung Lin
4,47841242
4,47841242
asked Feb 1 at 21:57
temp wattstemp watts
1126
asked Feb 1 at 21:57
temp wattstemp watts
1126
asked Feb 1 at 21:57
temp wattstemp watts
1126
1126
$begingroup$
I would try computing the first values, and then try to prove, using induction on $i$ a statement of the form "for all $igeq 5$, $f(p_i)geq ci$ for some constant $c$ (that you would somehow guess computing the first terms).
$endgroup$
– J.F
Feb 1 at 22:04
add a comment |
$begingroup$
I would try computing the first values, and then try to prove, using induction on $i$ a statement of the form "for all $igeq 5$, $f(p_i)geq ci$ for some constant $c$ (that you would somehow guess computing the first terms).
$endgroup$
– J.F
Feb 1 at 22:04
$begingroup$
I would try computing the first values, and then try to prove, using induction on $i$ a statement of the form "for all $igeq 5$, $f(p_i)geq ci$ for some constant $c$ (that you would somehow guess computing the first terms).
$endgroup$
– J.F
Feb 1 at 22:04
$begingroup$
I would try computing the first values, and then try to prove, using induction on $i$ a statement of the form "for all $igeq 5$, $f(p_i)geq ci$ for some constant $c$ (that you would somehow guess computing the first terms).
$endgroup$
– J.F
Feb 1 at 22:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The given recurrence equations are
begin{align}
g(p_i) &= frac{p_i - 2}{p_i} g(p_{i-1}) tag{1} \
f(p_i) &= frac{ p_i}2 g(p_i) tag{2}
end{align}
with the initial condition $displaystyle g(5) = frac35$.
Plug Eq.(1) into Eq.(2) we have at $i$ and $i+1$
begin{align}
f(p_i) &= frac{ p_i }2 g(p_i) = frac{ p_i - 2 }2 g( p_{i - 1} ) tag{3} \
f(p_{i+1}) &= frac{ p_{i+1} }2 g(p_{i+1}) = frac{ p_{i+1} - 2 }2 g( p_i ) tag{4}
end{align}
At the same time, reversing Eq.(2) one has $displaystyle g(p_i) = frac2{ p_i } f(p_i) $. Substitute this into Eq.(4) to obtain
$$ f(p_{i+1}) = frac{ p_{i+1} - 2 }{ p_i } f(p_i) = frac{ 7 + 4i }{ 5 + 4i } f(p_i) $$
since by definition $p_i = 5+4i$ and $p_{i+1} = 9 + 4i$.
The multiplication factor $displaystyle frac{ 7 + 4i }{ 5 + 4i } $ is strictly larger than unity while the initial $f(5) = frac32 > 0$ therefore $f( p_i ) to infty$ as $i to infty$.
answered Feb 3 at 3:34
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
$begingroup$
Thank you for your response. However, (7+4i)/(5+4i) approaches 1 as i goes to infinity. This may mean that the function can converge to a constant rather than go to infinity. Can you provide a more rigorous proof?
$endgroup$
– temp watts
Feb 3 at 16:49
$begingroup$
The situation you're worrying does not happen here. The textbooks by Rudin and Apostol are rigorous enough by most standards. See the cited restuls in wiki. In particular, the bullet below the line "if $r>1$, the series diverges" .
$endgroup$
– Lee David Chung Lin
Feb 4 at 7:35
$begingroup$
The wiki page you refer to says it is inconclusive since L=1 as n-> infinity. "This illustrates that when L = 1, the series may converge or diverge, and hence the original ratio test is inconclusive. In such cases, more refined tests are required to determine convergence or divergence. "
$endgroup$
– temp watts
Feb 4 at 17:18
$begingroup$
You'er not reading the part I was referring to. Please read my previous comment again carefully and don't just click on the link and randomly look for what you recognize. You should keep reading a few lines further down those you quoted in your comment.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:15
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3096773%2fprove-a-recursive-function-fp-i-p-i-2gp-i-to-infty-as-p-i-to-inft%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The given recurrence equations are
begin{align}
g(p_i) &= frac{p_i - 2}{p_i} g(p_{i-1}) tag{1} \
f(p_i) &= frac{ p_i}2 g(p_i) tag{2}
end{align}
with the initial condition $displaystyle g(5) = frac35$.
Plug Eq.(1) into Eq.(2) we have at $i$ and $i+1$
begin{align}
f(p_i) &= frac{ p_i }2 g(p_i) = frac{ p_i - 2 }2 g( p_{i - 1} ) tag{3} \
f(p_{i+1}) &= frac{ p_{i+1} }2 g(p_{i+1}) = frac{ p_{i+1} - 2 }2 g( p_i ) tag{4}
end{align}
At the same time, reversing Eq.(2) one has $displaystyle g(p_i) = frac2{ p_i } f(p_i) $. Substitute this into Eq.(4) to obtain
$$ f(p_{i+1}) = frac{ p_{i+1} - 2 }{ p_i } f(p_i) = frac{ 7 + 4i }{ 5 + 4i } f(p_i) $$
since by definition $p_i = 5+4i$ and $p_{i+1} = 9 + 4i$.
The multiplication factor $displaystyle frac{ 7 + 4i }{ 5 + 4i } $ is strictly larger than unity while the initial $f(5) = frac32 > 0$ therefore $f( p_i ) to infty$ as $i to infty$.
answered Feb 3 at 3:34
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
$begingroup$
Thank you for your response. However, (7+4i)/(5+4i) approaches 1 as i goes to infinity. This may mean that the function can converge to a constant rather than go to infinity. Can you provide a more rigorous proof?
$endgroup$
– temp watts
Feb 3 at 16:49
$begingroup$
The situation you're worrying does not happen here. The textbooks by Rudin and Apostol are rigorous enough by most standards. See the cited restuls in wiki. In particular, the bullet below the line "if $r>1$, the series diverges" .
$endgroup$
– Lee David Chung Lin
Feb 4 at 7:35
$begingroup$
The wiki page you refer to says it is inconclusive since L=1 as n-> infinity. "This illustrates that when L = 1, the series may converge or diverge, and hence the original ratio test is inconclusive. In such cases, more refined tests are required to determine convergence or divergence. "
$endgroup$
– temp watts
Feb 4 at 17:18
$begingroup$
You'er not reading the part I was referring to. Please read my previous comment again carefully and don't just click on the link and randomly look for what you recognize. You should keep reading a few lines further down those you quoted in your comment.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:15
add a comment |
$begingroup$
The given recurrence equations are
begin{align}
g(p_i) &= frac{p_i - 2}{p_i} g(p_{i-1}) tag{1} \
f(p_i) &= frac{ p_i}2 g(p_i) tag{2}
end{align}
with the initial condition $displaystyle g(5) = frac35$.
Plug Eq.(1) into Eq.(2) we have at $i$ and $i+1$
begin{align}
f(p_i) &= frac{ p_i }2 g(p_i) = frac{ p_i - 2 }2 g( p_{i - 1} ) tag{3} \
f(p_{i+1}) &= frac{ p_{i+1} }2 g(p_{i+1}) = frac{ p_{i+1} - 2 }2 g( p_i ) tag{4}
end{align}
At the same time, reversing Eq.(2) one has $displaystyle g(p_i) = frac2{ p_i } f(p_i) $. Substitute this into Eq.(4) to obtain
$$ f(p_{i+1}) = frac{ p_{i+1} - 2 }{ p_i } f(p_i) = frac{ 7 + 4i }{ 5 + 4i } f(p_i) $$
since by definition $p_i = 5+4i$ and $p_{i+1} = 9 + 4i$.
The multiplication factor $displaystyle frac{ 7 + 4i }{ 5 + 4i } $ is strictly larger than unity while the initial $f(5) = frac32 > 0$ therefore $f( p_i ) to infty$ as $i to infty$.
answered Feb 3 at 3:34
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
$begingroup$
Thank you for your response. However, (7+4i)/(5+4i) approaches 1 as i goes to infinity. This may mean that the function can converge to a constant rather than go to infinity. Can you provide a more rigorous proof?
$endgroup$
– temp watts
Feb 3 at 16:49
$begingroup$
The situation you're worrying does not happen here. The textbooks by Rudin and Apostol are rigorous enough by most standards. See the cited restuls in wiki. In particular, the bullet below the line "if $r>1$, the series diverges" .
$endgroup$
– Lee David Chung Lin
Feb 4 at 7:35
$begingroup$
The wiki page you refer to says it is inconclusive since L=1 as n-> infinity. "This illustrates that when L = 1, the series may converge or diverge, and hence the original ratio test is inconclusive. In such cases, more refined tests are required to determine convergence or divergence. "
$endgroup$
– temp watts
Feb 4 at 17:18
$begingroup$
You'er not reading the part I was referring to. Please read my previous comment again carefully and don't just click on the link and randomly look for what you recognize. You should keep reading a few lines further down those you quoted in your comment.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:15
add a comment |
$begingroup$
The given recurrence equations are
begin{align}
g(p_i) &= frac{p_i - 2}{p_i} g(p_{i-1}) tag{1} \
f(p_i) &= frac{ p_i}2 g(p_i) tag{2}
end{align}
with the initial condition $displaystyle g(5) = frac35$.
Plug Eq.(1) into Eq.(2) we have at $i$ and $i+1$
begin{align}
f(p_i) &= frac{ p_i }2 g(p_i) = frac{ p_i - 2 }2 g( p_{i - 1} ) tag{3} \
f(p_{i+1}) &= frac{ p_{i+1} }2 g(p_{i+1}) = frac{ p_{i+1} - 2 }2 g( p_i ) tag{4}
end{align}
At the same time, reversing Eq.(2) one has $displaystyle g(p_i) = frac2{ p_i } f(p_i) $. Substitute this into Eq.(4) to obtain
$$ f(p_{i+1}) = frac{ p_{i+1} - 2 }{ p_i } f(p_i) = frac{ 7 + 4i }{ 5 + 4i } f(p_i) $$
since by definition $p_i = 5+4i$ and $p_{i+1} = 9 + 4i$.
The multiplication factor $displaystyle frac{ 7 + 4i }{ 5 + 4i } $ is strictly larger than unity while the initial $f(5) = frac32 > 0$ therefore $f( p_i ) to infty$ as $i to infty$.
answered Feb 3 at 3:34
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
The given recurrence equations are
begin{align}
g(p_i) &= frac{p_i - 2}{p_i} g(p_{i-1}) tag{1} \
f(p_i) &= frac{ p_i}2 g(p_i) tag{2}
end{align}
with the initial condition $displaystyle g(5) = frac35$.
Plug Eq.(1) into Eq.(2) we have at $i$ and $i+1$
begin{align}
f(p_i) &= frac{ p_i }2 g(p_i) = frac{ p_i - 2 }2 g( p_{i - 1} ) tag{3} \
f(p_{i+1}) &= frac{ p_{i+1} }2 g(p_{i+1}) = frac{ p_{i+1} - 2 }2 g( p_i ) tag{4}
end{align}
At the same time, reversing Eq.(2) one has $displaystyle g(p_i) = frac2{ p_i } f(p_i) $. Substitute this into Eq.(4) to obtain
$$ f(p_{i+1}) = frac{ p_{i+1} - 2 }{ p_i } f(p_i) = frac{ 7 + 4i }{ 5 + 4i } f(p_i) $$
since by definition $p_i = 5+4i$ and $p_{i+1} = 9 + 4i$.
The multiplication factor $displaystyle frac{ 7 + 4i }{ 5 + 4i } $ is strictly larger than unity while the initial $f(5) = frac32 > 0$ therefore $f( p_i ) to infty$ as $i to infty$.
answered Feb 3 at 3:34
Lee David Chung LinLee David Chung Lin
4,47841242
answered Feb 3 at 3:34
Lee David Chung LinLee David Chung Lin
4,47841242
answered Feb 3 at 3:34
Lee David Chung LinLee David Chung Lin
4,47841242
answered Feb 3 at 3:34
Lee David Chung LinLee David Chung Lin
4,47841242
4,47841242
$begingroup$
Thank you for your response. However, (7+4i)/(5+4i) approaches 1 as i goes to infinity. This may mean that the function can converge to a constant rather than go to infinity. Can you provide a more rigorous proof?
$endgroup$
– temp watts
Feb 3 at 16:49
$begingroup$
The situation you're worrying does not happen here. The textbooks by Rudin and Apostol are rigorous enough by most standards. See the cited restuls in wiki. In particular, the bullet below the line "if $r>1$, the series diverges" .
$endgroup$
– Lee David Chung Lin
Feb 4 at 7:35
$begingroup$
The wiki page you refer to says it is inconclusive since L=1 as n-> infinity. "This illustrates that when L = 1, the series may converge or diverge, and hence the original ratio test is inconclusive. In such cases, more refined tests are required to determine convergence or divergence. "
$endgroup$
– temp watts
Feb 4 at 17:18
$begingroup$
You'er not reading the part I was referring to. Please read my previous comment again carefully and don't just click on the link and randomly look for what you recognize. You should keep reading a few lines further down those you quoted in your comment.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:15
add a comment |
$begingroup$
Thank you for your response. However, (7+4i)/(5+4i) approaches 1 as i goes to infinity. This may mean that the function can converge to a constant rather than go to infinity. Can you provide a more rigorous proof?
$endgroup$
– temp watts
Feb 3 at 16:49
$begingroup$
The situation you're worrying does not happen here. The textbooks by Rudin and Apostol are rigorous enough by most standards. See the cited restuls in wiki. In particular, the bullet below the line "if $r>1$, the series diverges" .
$endgroup$
– Lee David Chung Lin
Feb 4 at 7:35
$begingroup$
The wiki page you refer to says it is inconclusive since L=1 as n-> infinity. "This illustrates that when L = 1, the series may converge or diverge, and hence the original ratio test is inconclusive. In such cases, more refined tests are required to determine convergence or divergence. "
$endgroup$
– temp watts
Feb 4 at 17:18
$begingroup$
You'er not reading the part I was referring to. Please read my previous comment again carefully and don't just click on the link and randomly look for what you recognize. You should keep reading a few lines further down those you quoted in your comment.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:15
$begingroup$
Thank you for your response. However, (7+4i)/(5+4i) approaches 1 as i goes to infinity. This may mean that the function can converge to a constant rather than go to infinity. Can you provide a more rigorous proof?
$endgroup$
– temp watts
Feb 3 at 16:49
$begingroup$
Thank you for your response. However, (7+4i)/(5+4i) approaches 1 as i goes to infinity. This may mean that the function can converge to a constant rather than go to infinity. Can you provide a more rigorous proof?
$endgroup$
– temp watts
Feb 3 at 16:49
$begingroup$
The situation you're worrying does not happen here. The textbooks by Rudin and Apostol are rigorous enough by most standards. See the cited restuls in wiki. In particular, the bullet below the line "if $r>1$, the series diverges" .
$endgroup$
– Lee David Chung Lin
Feb 4 at 7:35
$begingroup$
The situation you're worrying does not happen here. The textbooks by Rudin and Apostol are rigorous enough by most standards. See the cited restuls in wiki. In particular, the bullet below the line "if $r>1$, the series diverges" .
$endgroup$
– Lee David Chung Lin
Feb 4 at 7:35
$begingroup$
The wiki page you refer to says it is inconclusive since L=1 as n-> infinity. "This illustrates that when L = 1, the series may converge or diverge, and hence the original ratio test is inconclusive. In such cases, more refined tests are required to determine convergence or divergence. "
$endgroup$
– temp watts
Feb 4 at 17:18
$begingroup$
The wiki page you refer to says it is inconclusive since L=1 as n-> infinity. "This illustrates that when L = 1, the series may converge or diverge, and hence the original ratio test is inconclusive. In such cases, more refined tests are required to determine convergence or divergence. "
$endgroup$
– temp watts
Feb 4 at 17:18
$begingroup$
You'er not reading the part I was referring to. Please read my previous comment again carefully and don't just click on the link and randomly look for what you recognize. You should keep reading a few lines further down those you quoted in your comment.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:15
$begingroup$
You'er not reading the part I was referring to. Please read my previous comment again carefully and don't just click on the link and randomly look for what you recognize. You should keep reading a few lines further down those you quoted in your comment.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:15
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3096773%2fprove-a-recursive-function-fp-i-p-i-2gp-i-to-infty-as-p-i-to-inft%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I would try computing the first values, and then try to prove, using induction on $i$ a statement of the form "for all $igeq 5$, $f(p_i)geq ci$ for some constant $c$ (that you would somehow guess computing the first terms).
$endgroup$
– J.F
Feb 1 at 22:04