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Determine the value for x^5 + 1/x^5

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anemone

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Feb 14, 2012
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If \(\displaystyle x^2+\frac{1}{x^2}=7\), find all possible values of \(\displaystyle x^5+\frac{1}{x^5}\).
 

Prove It

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MHB Math Helper
Jan 26, 2012
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Re: Determine the value for x^5+1/x^5

If \(\displaystyle x^2+\frac{1}{x^2}=7\), find all possible values of \(\displaystyle x^5+\frac{1}{x^5}\).
[tex]\displaystyle \begin{align*} x^2 + \frac{1}{x^2} &= 7 \\ x^2 \left( x^2 + \frac{1}{x^2} \right) &= 7x^2 \\ x^4 + 1 &= 7x^2 \\ x^4 - 7x^2 + 1 &= 0 \\ X^2 - 7X + 1 &= 0 \textrm{ if }X = x^2 \\ X^2 - 7X + \left( -\frac{7}{2} \right) ^2 - \left( -\frac{7}{2} \right) ^2 + 1 &= 0 \\ \left( X - \frac{7}{2} \right) ^2 - \frac{49}{4} + \frac{4}{4} &= 0 \\ \left( X - \frac{7}{2} \right) ^2 - \frac{45}{4} &= 0 \\ \left( X - \frac{7}{2} \right) ^2 &= \frac{45}{4} \\ X - \frac{7}{2} &= \pm \frac{ 3\sqrt{5}}{2} \\ X &= \frac{7 \pm 3\sqrt{5}}{2} \\ x^2 &= \frac{7 \pm 3\sqrt{5}}{2} \\ x &= \pm \sqrt{ \frac{7 \pm 3\sqrt{5}}{2} } \\ x &= \pm \frac{\sqrt{14 \pm 6\sqrt{5}}}{2} \end{align*} [/tex]

so

[tex]\displaystyle \begin{align*} x^5 + \frac{1}{x^5} &= \left( \pm \frac{ \sqrt{ 14 \pm 6\sqrt{5} } }{2} \right) ^5 + \frac{1}{ \left( \pm \frac{ \sqrt{ 14 \pm 6\sqrt{5} } }{2} \right) ^5 } \end{align*}[/tex]

which probably simplifies quite a lot :)
 

MarkFL

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Feb 24, 2012
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Re: Determine the value for x^5+1/x^5

My solution:

\(\displaystyle \left(x+\frac{1}{x} \right)^2=x^2+2+\frac{1}{x^2}=9\)

Hence:

\(\displaystyle x+\frac{1}{x}=\pm3\)

Also, we find:

\(\displaystyle \left(x+\frac{1}{x} \right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x} \right)\)

\(\displaystyle \pm27=x^3+\frac{1}{x^3}\pm9\)

\(\displaystyle x^3+\frac{1}{x^3}=\pm18\)

And finally:

\(\displaystyle \left(x+\frac{1}{x} \right)^5=x^5+\frac{1}{x^5}+5\left(x^3+\frac{1}{x^3} \right)+10\left(x+\frac{1}{x} \right)\)

Hence:

\(\displaystyle x^5+\frac{1}{x^5}=\pm(243-5\cdot18-10\cdot3)=\pm123\)
 
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anemone

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Feb 14, 2012
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Re: Determine the value for x^5+1/x^5

[tex]\displaystyle \begin{align*} x^2 + \frac{1}{x^2} &= 7 \\ x^2 \left( x^2 + \frac{1}{x^2} \right) &= 7x^2 \\ x^4 + 1 &= 7x^2 \\ x^4 - 7x^2 + 1 &= 0 \\ X^2 - 7X + 1 &= 0 \textrm{ if }X = x^2 \\ X^2 - 7X + \left( -\frac{7}{2} \right) ^2 - \left( -\frac{7}{2} \right) ^2 + 1 &= 0 \\ \left( X - \frac{7}{2} \right) ^2 - \frac{49}{4} + \frac{4}{4} &= 0 \\ \left( X - \frac{7}{2} \right) ^2 - \frac{45}{4} &= 0 \\ \left( X - \frac{7}{2} \right) ^2 &= \frac{45}{4} \\ X - \frac{7}{2} &= \pm \frac{ 3\sqrt{5}}{2} \\ X &= \frac{7 \pm 3\sqrt{5}}{2} \\ x^2 &= \frac{7 \pm 3\sqrt{5}}{2} \\ x &= \pm \sqrt{ \frac{7 \pm 3\sqrt{5}}{2} } \\ x &= \pm \frac{\sqrt{14 \pm 6\sqrt{5}}}{2} \end{align*} [/tex]

so

[tex]\displaystyle \begin{align*} x^5 + \frac{1}{x^5} &= \left( \pm \frac{ \sqrt{ 14 \pm 6\sqrt{5} } }{2} \right) ^5 + \frac{1}{ \left( \pm \frac{ \sqrt{ 14 \pm 6\sqrt{5} } }{2} \right) ^5 } \end{align*}[/tex]

which probably simplifies quite a lot :)
Hi Prove It,

After reading your step-by-step solution, I really think your answer is correct but just it needed more simplification to it! :eek:And thanks for participating!

My solution:

\(\displaystyle \left(x+\frac{1}{x} \right)^2=x^2+2+\frac{1}{x^2}=9\)

Hence:

\(\displaystyle x+\frac{1}{x}=\pm3\)

Also, we find:

\(\displaystyle \left(x+\frac{1}{x} \right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x} \right)\)

\(\displaystyle \pm27=x^3+\frac{1}{x^3}\pm9\)

\(\displaystyle x^3+\frac{1}{x^3}=\pm18\)

And finally:

\(\displaystyle \left(x+\frac{1}{x} \right)^5=x^5+\frac{1}{x^5}+5\left(x^3+\frac{1}{x^3} \right)+10\left(x+\frac{1}{x} \right)\)

Hence:

\(\displaystyle x^5+\frac{1}{x^5}=\pm(243-5\cdot18-10\cdot3)=\pm123\)
Well done, MarkFL! Your answer is correct and your method of approaching it is different from mine as I used trigonometric method to solve it and it's more tedious and lousy than yours!(Nerd)
 

MarkFL

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Feb 24, 2012
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Re: Determine the value for x^5+1/x^5

I still would like to see it! (Clapping)
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
Re: Determine the value for x^5+1/x^5

Good solution by MarkFL

Here is another

Let f(n) = x^n + 1/x^n = x^n + x^(-n) note it should be strictly f(n,x) but I have taken x implicitly)
Then f(0) = 2
Now
f(n)f(m) = (x^n + x^(-n)) ( x^m + x^(-m) = x^(n+m) + x^(n-m) + x^-(n-m) + x^(-(n+m))
= x^(n+m) + x^(-(n+m)) + x^(n-m) + x^-(n-m)
= f(n+m) + f(n-m)

so we got f(n)f(m) = f(n+m) + f(n-m)

n= 1,m= 1
give f(2) + f(0) = f(1)^2 = 9 so f(1) = 3 or – 3

for f(1) = 3
now n = 2, m= 1 we get

f(2)f(1) = f(3) + f(1) or f(3) = f(2) f(1) – f(1) = 7 * 3 – 3 = 18

n =3 , m= 2 give

f(3)f(2) = f(5) + f(1) or f(5) = f(3) f(2) – f(1) = 18 * 7 – 3 = 126 -3 = 123


for f(1) = - 3 we can compute similarly
 
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anemone

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Feb 14, 2012
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Re: Determine the value for x^5+1/x^5

Good solution by MarkFL

Here is another

Let f(n) = x^n + 1/x^n = x^n + x^(-n) note it should be strictly f(n,x) but I have taken x implicitly)
Then f(0) = 2
Now
f(n)f(m) = (x^n + x^(-n)) ( x^m + x^(-m) = x^(n+m) + x^(n-m) + x^-(n-m) + x^(-(n+m))
= x^(n+m) + x^(-(n+m)) + x^(n-m) + x^-(n-m)
= f(n+m) + f(n-m)

so we got f(n)f(m) = f(n+m) + f(n-m)

n= 1,m= 1
give f(2) + f(0) = f(1)^2 = 9 so f(1) = 3 or – 3

for f(1) = 3
now n = 2, m= 1 we get

f(2)f(1) = f(3) + f(1) or f(3) = f(2) f(1) – f(1) = 7 * 3 – 3 = 18

n =3 , m= 2 give

f(3)f(2) = f(5) + f(1) or f(5) = f(3) f(2) – f(1) = 18 * 7 – 3 = 126 -3 = 123


for f(1) = - 3 we can compute similarly
Hi kaliprasad,

Thanks for participating and your method has given me a completely new insight on how to tackle the problem using another route...well done, kali!(Sun)
 
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anemone

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Feb 14, 2012
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Re: Determine the value for x^5+1/x^5

I still would like to see it! (Clapping)
My solution:

First I let $x^2=\tan y$, then \(\displaystyle x^2+\frac{1}{x^2}=7\) becomes \(\displaystyle \tan y+\frac{1}{\tan y}=7\) or \(\displaystyle \tan^2 y+1=7\tan y\) and this implies

i.) \(\displaystyle \sec^2 y=7\tan y\)

ii.)\(\displaystyle \sin 2y=\frac{2}{7}\),

iii.) \(\displaystyle \sin y \cos y=\frac{1}{7}\),

iv.) \(\displaystyle \sin y+\cos y=\frac{3}{\sqrt{7}}\),

v.) \(\displaystyle \sin^4 y+\cos^4 y=\frac{47}{49}\),

vi.) \(\displaystyle \sqrt{\tan y}\cos y= \pm \frac{1}{\sqrt{7}}\),

Next, we have to evaluate \(\displaystyle x^5+\frac{1}{x^5}\), i.e.

\(\displaystyle x^5+\frac{1}{x^5}=\frac{x^{10}+1}{x^5}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{(x^{2})^5+1}{(x^2)^4(x)}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{(\tan y)^5+1}{(\tan y)^4(\sqrt{\tan y})}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{\sin^5 y+\cos^5 y}{\cos^3 y \sin^2 y(\sqrt{\tan y})}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{(\sin y+\cos y)(\sin^4 y+\cos^4 y-\sin y \cos y(\sin^2 y + \cos^2 y- \sin y \cos y))}{(\cos y \sin y)^2(\sqrt{\tan y})\cos y}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\pm \frac{(\frac{3}{\sqrt{7}})(\frac{47}{49}-\frac{1}{7}(1- \frac{1}{7}))}{(\frac{1}{7})^2(\frac{1}{\sqrt{7}})}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=\pm 123\)


I know, I know, this looks like a mess:mad:...but this is my thought and I must post it to show how much I appreciate to the fact that there are so many members answered to my challenge problems!:)
 
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MarkFL

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Feb 24, 2012
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Re: Determine the value for x^5+1/x^5

I just want to say it speaks highly of your character to be willing to post your solution, even if you feel it is not as straightforward as that given by another. But, I think others benefit by seeing different approaches, so you have done a good thing for MHB. (Inlove)

I know I have posted solutions to problems, only to have someone else come along and post a much more direct and simple solution, and the feeling of "D'oh! Why didn't I think of that?" can momentarily be very powerful! (Doh)

But, in the end, we can console ourselves with the knowledge that at least we were able to solve the problem, and we can learn from the solutions given by others. (Sun)

Thank you for sharing your solution. (Clapping)
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Re: Determine the value for x^5+1/x^5

I just want to say it speaks highly of your character to be willing to post your solution, even if you feel it is not as straightforward as that given by another. But, I think others benefit by seeing different approaches, so you have done a good thing for MHB. (Inlove)

I know I have posted solutions to problems, only to have someone else come along and post a much more direct and simple solution, and the feeling of "D'oh! Why didn't I think of that?" can momentarily be very powerful! (Doh)

But, in the end, we can console ourselves with the knowledge that at least we were able to solve the problem, and we can learn from the solutions given by others. (Sun)

Thank you for sharing your solution. (Clapping)
Thank you for your kind words, Mark! I agree with all that you said and you're always a joy to read!
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Re: Determine the value for x^5+1/x^5

Extremely well said, Mark^^ (Clapping)


For me personally, when an problem I find interesting gets posted on here - which thankfully happens a lot - I enjoy looking at all of the different solutions. And sometimes, like in this case, I find the less obvious / more circumspect ones the most interesting...

Well done Anemone! And thanks for sharing!! (Sun)
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Re: Determine the value for x^5+1/x^5

Extremely well said, Mark^^ (Clapping)


For me personally, when an problem I find interesting gets posted on here - which thankfully happens a lot - I enjoy looking at all of the different solutions. And sometimes, like in this case, I find the less obvious / more circumspect ones the most interesting...

Well done Anemone! And thanks for sharing!! (Sun)
Hi Dreamweaver,

That's extremely nice of you to say...and as a matter of fact, I know that I am the one who owes thanks to those members who have taught me how to solve the math problems on which I was stuck, and to those who have participated in my challenge problems, and to those who have read the challenge problems that I posted! :)