determining if a differential equation has unique solution

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
5
down vote

favorite
1












Our teacher asked a relatively simple question




What is the value of $f(2)$ if $f(f(x))=16x-15?$




I found a solution assuming $f(x)$ was in the form of $ax+b$, but how do I show that $f(x)$ must be in that form?

I thought of taking derivative of both sides and reached
$$f'(f(x))=16/f'(x)$$
How can I show that this function has a unique solution or it doesn't have a unique solution?






differential-equations







share|cite|improve this question



















  • 1




    For future reference: math.meta.stackexchange.com/questions/5020/…
    – 5xum
    5 hours ago














up vote
5
down vote

favorite
1












Our teacher asked a relatively simple question




What is the value of $f(2)$ if $f(f(x))=16x-15?$




I found a solution assuming $f(x)$ was in the form of $ax+b$, but how do I show that $f(x)$ must be in that form?

I thought of taking derivative of both sides and reached
$$f'(f(x))=16/f'(x)$$
How can I show that this function has a unique solution or it doesn't have a unique solution?






differential-equations







share|cite|improve this question



















  • 1




    For future reference: math.meta.stackexchange.com/questions/5020/…
    – 5xum
    5 hours ago












up vote
5
down vote

favorite
1









up vote
5
down vote

favorite
1






1





Our teacher asked a relatively simple question




What is the value of $f(2)$ if $f(f(x))=16x-15?$




I found a solution assuming $f(x)$ was in the form of $ax+b$, but how do I show that $f(x)$ must be in that form?

I thought of taking derivative of both sides and reached
$$f'(f(x))=16/f'(x)$$
How can I show that this function has a unique solution or it doesn't have a unique solution?






differential-equations







share|cite|improve this question















Our teacher asked a relatively simple question




What is the value of $f(2)$ if $f(f(x))=16x-15?$




I found a solution assuming $f(x)$ was in the form of $ax+b$, but how do I show that $f(x)$ must be in that form?

I thought of taking derivative of both sides and reached
$$f'(f(x))=16/f'(x)$$
How can I show that this function has a unique solution or it doesn't have a unique solution?






differential-equations




differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









humanStampedist

2,148213




2,148213










asked 5 hours ago









alper akyuz

309111




309111







  • 1




    For future reference: math.meta.stackexchange.com/questions/5020/…
    – 5xum
    5 hours ago












  • 1




    For future reference: math.meta.stackexchange.com/questions/5020/…
    – 5xum
    5 hours ago







1




1




For future reference: math.meta.stackexchange.com/questions/5020/…
– 5xum
5 hours ago




For future reference: math.meta.stackexchange.com/questions/5020/…
– 5xum
5 hours ago










1 Answer
1






active

oldest

votes

















up vote
5
down vote



accepted










Apply $f$ from the inside and outside, set $x=f(w)$ then
$$
f(f(f(w)))=16f(w)-15
$$
and also
$$
f(f(f(x)))=f(16x-15)
$$
so that $f(16x-15)=16f(x)-15$, from which one concludes directly $f(1)=1$. Set $g(x)=f(1+x)-1$, then $16g(x)=f(1+16x)-1=g(16x)$, $g(0)=0$ or equivalently
$$
g(x)=16,gbiggl(fracx16biggr)=16^n,gbiggl(fracx16^nbiggr)~~forall ninBbb N
$$
Assuming differentiability of $f$ and thus of $g$ one concludes that $g(x)=ax$, $f(x)=ax+1-a$, validating your first guess as the only solution,
$$
f(f(x))=a(ax+1-a)+1-a=a^2x+1-a^2implies a=pm4, ~~f(2)=a+1in-3,5.
$$






share|cite|improve this answer




















  • I used a much more complicated method to arrive at the same answer but your method is much more elegant.
    – alper akyuz
    3 hours ago










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',
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
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2986911%2fdetermining-if-a-differential-equation-has-unique-solution%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










Apply $f$ from the inside and outside, set $x=f(w)$ then
$$
f(f(f(w)))=16f(w)-15
$$
and also
$$
f(f(f(x)))=f(16x-15)
$$
so that $f(16x-15)=16f(x)-15$, from which one concludes directly $f(1)=1$. Set $g(x)=f(1+x)-1$, then $16g(x)=f(1+16x)-1=g(16x)$, $g(0)=0$ or equivalently
$$
g(x)=16,gbiggl(fracx16biggr)=16^n,gbiggl(fracx16^nbiggr)~~forall ninBbb N
$$
Assuming differentiability of $f$ and thus of $g$ one concludes that $g(x)=ax$, $f(x)=ax+1-a$, validating your first guess as the only solution,
$$
f(f(x))=a(ax+1-a)+1-a=a^2x+1-a^2implies a=pm4, ~~f(2)=a+1in-3,5.
$$






share|cite|improve this answer




















  • I used a much more complicated method to arrive at the same answer but your method is much more elegant.
    – alper akyuz
    3 hours ago














up vote
5
down vote



accepted










Apply $f$ from the inside and outside, set $x=f(w)$ then
$$
f(f(f(w)))=16f(w)-15
$$
and also
$$
f(f(f(x)))=f(16x-15)
$$
so that $f(16x-15)=16f(x)-15$, from which one concludes directly $f(1)=1$. Set $g(x)=f(1+x)-1$, then $16g(x)=f(1+16x)-1=g(16x)$, $g(0)=0$ or equivalently
$$
g(x)=16,gbiggl(fracx16biggr)=16^n,gbiggl(fracx16^nbiggr)~~forall ninBbb N
$$
Assuming differentiability of $f$ and thus of $g$ one concludes that $g(x)=ax$, $f(x)=ax+1-a$, validating your first guess as the only solution,
$$
f(f(x))=a(ax+1-a)+1-a=a^2x+1-a^2implies a=pm4, ~~f(2)=a+1in-3,5.
$$






share|cite|improve this answer




















  • I used a much more complicated method to arrive at the same answer but your method is much more elegant.
    – alper akyuz
    3 hours ago












up vote
5
down vote



accepted







up vote
5
down vote



accepted






Apply $f$ from the inside and outside, set $x=f(w)$ then
$$
f(f(f(w)))=16f(w)-15
$$
and also
$$
f(f(f(x)))=f(16x-15)
$$
so that $f(16x-15)=16f(x)-15$, from which one concludes directly $f(1)=1$. Set $g(x)=f(1+x)-1$, then $16g(x)=f(1+16x)-1=g(16x)$, $g(0)=0$ or equivalently
$$
g(x)=16,gbiggl(fracx16biggr)=16^n,gbiggl(fracx16^nbiggr)~~forall ninBbb N
$$
Assuming differentiability of $f$ and thus of $g$ one concludes that $g(x)=ax$, $f(x)=ax+1-a$, validating your first guess as the only solution,
$$
f(f(x))=a(ax+1-a)+1-a=a^2x+1-a^2implies a=pm4, ~~f(2)=a+1in-3,5.
$$






share|cite|improve this answer












Apply $f$ from the inside and outside, set $x=f(w)$ then
$$
f(f(f(w)))=16f(w)-15
$$
and also
$$
f(f(f(x)))=f(16x-15)
$$
so that $f(16x-15)=16f(x)-15$, from which one concludes directly $f(1)=1$. Set $g(x)=f(1+x)-1$, then $16g(x)=f(1+16x)-1=g(16x)$, $g(0)=0$ or equivalently
$$
g(x)=16,gbiggl(fracx16biggr)=16^n,gbiggl(fracx16^nbiggr)~~forall ninBbb N
$$
Assuming differentiability of $f$ and thus of $g$ one concludes that $g(x)=ax$, $f(x)=ax+1-a$, validating your first guess as the only solution,
$$
f(f(x))=a(ax+1-a)+1-a=a^2x+1-a^2implies a=pm4, ~~f(2)=a+1in-3,5.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









LutzL

52.3k41851




52.3k41851











  • I used a much more complicated method to arrive at the same answer but your method is much more elegant.
    – alper akyuz
    3 hours ago
















  • I used a much more complicated method to arrive at the same answer but your method is much more elegant.
    – alper akyuz
    3 hours ago















I used a much more complicated method to arrive at the same answer but your method is much more elegant.
– alper akyuz
3 hours ago




I used a much more complicated method to arrive at the same answer but your method is much more elegant.
– alper akyuz
3 hours ago

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2986911%2fdetermining-if-a-differential-equation-has-unique-solution%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What does second last employer means? [closed]

Image
Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
Read more

List of Gilmore Girls characters

Image
The season three cast—(Back row): Lane, Michel, Paris, Emily, Richard, Sookie, Miss Patty, Kirk; Front row: Jess, Luke, Lorelai, Rory, Dean This is a list of characters for the comedy-drama television series Gilmore Girls . Contents 1 Main characters 1.1 Lorelai Gilmore 1.2 Rory Gilmore 1.3 Sookie St. James 1.4 Lane Kim 1.5 Michel Gerard 1.6 Luke Danes 1.7 Emily Gilmore 1.8 Richard Gilmore 1.9 Paris Geller 1.10 Dean Forester 1.11 Jess Mariano 1.12 Kirk Gleason 1.13 Jason Stiles 1.14 Logan Huntzberger 2 Major recurring characters 2.1 Christopher Hayden 2.2 Jackson Belleville 2.3 Mrs. Kim 2.4 Tristin DuGray 2.5 Max Medina 2.6 Taylor Doose 2.7 Dave Rygalski 2.8 Marty 3 Recurring characters 3.1 Stars Hollow 3.2 Chilton 3.3 Yale 3.4 Other 4 Notable guest stars 5 Unseen characters 6 Notes 7 References Main characters Actor Character Appearances Season 1 Season 2 Season 3 Season 4 Season 5 Season 6 Season 7 A Year in the Life Lauren Graham Lorelai Gilmore...
Read more

Confectionery

Image
Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of ...
Read more