the equations are:
2(a^4) - λ (a^3) -2 = 0
2(b^4) - λ (b^3) -2 = 0
a+b-1=0
I can solve by trial and error to get a=b=0.5 but what if there is another root for these equations
Homework Statement
How to prove (a+1/a)^2 + (b+1/b)^2 >= 25/2 given that a+b=1 and a,b positive
Homework Equations
The Attempt at a Solution
I tried replacing b with 1-a but I get 6th degree a which I don't know how to find inequality for.
Since a and b are perfect symmetric in...
\sum_{r=k+1}^{n+1} \frac {1} {(r-1)!} - \sum_{r=k+1}^{n+1} \frac {1} {(r)!}
YOUR RIGHT! it becomes simply
\frac {1} {(k)!} - \frac {1} {(n+1)!}
Just 1 final question?
I take lim as n tends to infinity of my answer so I get that the infinite sum of all terms after ak/k! is...
\sum_{r=k+1}^{n+1} \frac {r-1} {(r)!} < \frac 1 {k!}
Is this correct?
\sum_{r=0}^{inf} \frac {r-1} {(r)!} = e-e=0
And then I take out term from 0 to k?
I think that your previous answer using series will solve it.
I am thinking of putting
\sum_{r=k}^n \frac r {(r+1)!} < \frac 1 {k!}
and I use the poisson distribution series to prove = 1/k! and thus the sum of any finite terms of the series is < 1/k!
Am I correct?
As I can see from the formula of cauchy inequality:
(a1^2+a2^2+...+an^2)^1/2 . (b1^2+b2^2+...+bn)^1/2 >= a1b1+a2b2 + ... + anbn
Can I conclude from the above formula that:
(a1+a2+...+an)^1/2 . (b1+b2+...+bn)^1/2 >= (a1b1)^1/2 + (a2b2)^1/2 +...+ (anbn)^1/2
by setting a1,...,an =...
This is not the question of my homework. I will write it here down to show you that I am doing an effort and what I am asking will lead me to the answer.
Question:
Prove every positive rational number x can be expressed in ONE way in the form
x= a1 + a2/2! + a3/3! + ... + ak/k!
where...
Yes the second one you wrote is what i exactly mean. How do I prove it? Also I think if i set n -> infinity (find lim at infinity), I would get = instead of < . Am I correct?