Proving $a_n>2$ where $a_k=sqrt{2a_{k-1}}$
$begingroup$
Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.
$a_1 = 3$.
Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.
discrete-mathematics
edited Nov 27 '18 at 4:42
Blue
48k870153
asked Nov 27 '18 at 2:20
capcap
103
$endgroup$
add a comment |
$begingroup$
Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.
$a_1 = 3$.
Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.
discrete-mathematics
edited Nov 27 '18 at 4:42
Blue
48k870153
asked Nov 27 '18 at 2:20
capcap
103
$endgroup$
3
$begingroup$
Stop trying to delete your questions.
$endgroup$
– fleablood
Nov 27 '18 at 3:12
$begingroup$
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
$endgroup$
– fleablood
Nov 27 '18 at 5:26
add a comment |
$begingroup$
Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.
$a_1 = 3$.
Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.
discrete-mathematics
edited Nov 27 '18 at 4:42
Blue
48k870153
asked Nov 27 '18 at 2:20
capcap
103
$endgroup$
Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.
$a_1 = 3$.
Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.
discrete-mathematics
discrete-mathematics
edited Nov 27 '18 at 4:42
Blue
48k870153
asked Nov 27 '18 at 2:20
capcap
103
edited Nov 27 '18 at 4:42
Blue
48k870153
asked Nov 27 '18 at 2:20
capcap
103
edited Nov 27 '18 at 4:42
Blue
48k870153
edited Nov 27 '18 at 4:42
Blue
48k870153
edited Nov 27 '18 at 4:42
Blue
48k870153
48k870153
asked Nov 27 '18 at 2:20
capcap
103
asked Nov 27 '18 at 2:20
capcap
103
asked Nov 27 '18 at 2:20
capcap
103
103
3
$begingroup$
Stop trying to delete your questions.
$endgroup$
– fleablood
Nov 27 '18 at 3:12
$begingroup$
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
$endgroup$
– fleablood
Nov 27 '18 at 5:26
add a comment |
3
$begingroup$
Stop trying to delete your questions.
$endgroup$
– fleablood
Nov 27 '18 at 3:12
$begingroup$
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
$endgroup$
– fleablood
Nov 27 '18 at 5:26
3
3
$begingroup$
Stop trying to delete your questions.
$endgroup$
– fleablood
Nov 27 '18 at 3:12
$begingroup$
Stop trying to delete your questions.
$endgroup$
– fleablood
Nov 27 '18 at 3:12
$begingroup$
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
$endgroup$
– fleablood
Nov 27 '18 at 5:26
$begingroup$
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
$endgroup$
– fleablood
Nov 27 '18 at 5:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You first need to show your base case. For $k = 2$, we have
$$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$
Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.
We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
begin{align}
a_{k+1} &= sqrt{2 a_k}\
&= (sqrt{2})(sqrt{a_k})\
end{align}
But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.
Hence by the principle of mathematical induction, we are done :)
answered Nov 27 '18 at 2:52
AlkaKadriAlkaKadri
1,459411
$endgroup$
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%2f3015244%2fproving-a-n2-where-a-k-sqrt2a-k-1%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
$begingroup$
You first need to show your base case. For $k = 2$, we have
$$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$
Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.
We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
begin{align}
a_{k+1} &= sqrt{2 a_k}\
&= (sqrt{2})(sqrt{a_k})\
end{align}
But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.
Hence by the principle of mathematical induction, we are done :)
answered Nov 27 '18 at 2:52
AlkaKadriAlkaKadri
1,459411
$endgroup$
add a comment |
$begingroup$
You first need to show your base case. For $k = 2$, we have
$$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$
Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.
We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
begin{align}
a_{k+1} &= sqrt{2 a_k}\
&= (sqrt{2})(sqrt{a_k})\
end{align}
But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.
Hence by the principle of mathematical induction, we are done :)
answered Nov 27 '18 at 2:52
AlkaKadriAlkaKadri
1,459411
$endgroup$
add a comment |
$begingroup$
You first need to show your base case. For $k = 2$, we have
$$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$
Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.
We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
begin{align}
a_{k+1} &= sqrt{2 a_k}\
&= (sqrt{2})(sqrt{a_k})\
end{align}
But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.
Hence by the principle of mathematical induction, we are done :)
answered Nov 27 '18 at 2:52
AlkaKadriAlkaKadri
1,459411
$endgroup$
You first need to show your base case. For $k = 2$, we have
$$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$
Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.
We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
begin{align}
a_{k+1} &= sqrt{2 a_k}\
&= (sqrt{2})(sqrt{a_k})\
end{align}
But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.
Hence by the principle of mathematical induction, we are done :)
answered Nov 27 '18 at 2:52
AlkaKadriAlkaKadri
1,459411
answered Nov 27 '18 at 2:52
AlkaKadriAlkaKadri
1,459411
answered Nov 27 '18 at 2:52
AlkaKadriAlkaKadri
1,459411
answered Nov 27 '18 at 2:52
AlkaKadriAlkaKadri
1,459411
1,459411
add a comment |
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%2f3015244%2fproving-a-n2-where-a-k-sqrt2a-k-1%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
3
$begingroup$
Stop trying to delete your questions.
$endgroup$
– fleablood
Nov 27 '18 at 3:12
$begingroup$
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
$endgroup$
– fleablood
Nov 27 '18 at 5:26