# Proof for Rieman Hypothesis

1. Apr 2, 2009

### imag94

$$Proof for Riemann Hypothesis Abstract: Proof of Riemann’s hypothesis that the real part of the solution of Zeta function is ½ is proved. Historical development of this area of Mathematics from Gauss, Legrange, Euler, Riemann to Hilbert is discussed. Initially a surrogate for zeta function is derived and using Cauchy principal number of integration between bounds it is proved that the real part of Riemann zeta function is ½. Also in the graph_3 the area between 2 and ½ is zero proving again the same. Then a differential equation with c=3 is derived and the solution to it’s reciprocal which is the zeta function again gives the same result the real part of the root as ½. Now c is derived differently from first principles as 0.25 and the real part of it’s root is found to be again ½. It’s graph which is Graph_8 is found to be similar to Graph_3 the surrogate. The different scales might make them look slightly different, one is using decimal numbers and one using trigonometric numbers. Finally the differential equation is taken which is primal of the reciprocals and it’s root is found to be 2 proving again ½ which is its reciprocal as the real part of the zeta function. Subscription: Legrange and Gauss conjured that п(x) the function counting all the primes less than x asymptotically approaches Li(x) meaning п(x)/Li(x) tend to 1, where, $$Li(x) =\int_2 ^n \frac {dx}{ln(x)} dx$$ Euler created a time series solution to the function Li(x) and Riemann named it the ξ function adding his own solution to Euler’s work. In Riemann’s words “a value x is the root of a function f(x) if f(x)=0. A root of the function ξ(x) is real if and only if the root of the zeta function is complex number with real part equal to ½”. Proving the real part to be ½ was left undone by Riemann. Hilbert later on added, finding the proof for Riemann hypothesis as one of the problem that remain un resolved in Mathematics. I start where Riemann left his hypothesis without the proof. As said earlier $$Li(x) =\int_2 ^n \frac {dx}{ln(x)} dx$$ Now, Assume $$a = \Ln x^x$$ $$x^x = e^{x.ln x}$$ $$a = x\ln (x)$$ $$\ln x =\frac {a}{x}$$ $$\xi (x) =\int _2 ^n \frac {x}{a} dx$$ = $$= \frac {x^2}{2a}$$ Now to get the Error estimate we add п/ln(x) to the above function. Then if we graph the zeta function $$\frac {x^2}{2a} + \frac {\pi}{ln x}$$ we see the root falling at exactly That is area under the graph between 2 and ½ is zero. If this integral is evaluated between 2 and ½ we will see that the zeta function going to 0. The integral is evaluated using Cauchy’s principal number 1, between limits 2 and 1 and then 1 and 1/2 That is, $$\frac {x^2}{2Ln x^x}_2 ^1 + \frac {x^2}{2Ln x^x}_1 ^1/2$$ = 1/(2. ln (1)) – 4/(2. ln(4)) + 0.707/(2.ln(0.707) – 1/(2.ln(1)) =0 This concludes the proof that Rieman’s Zeta function has it’s root as ½ +/- i0 since the area under the under the zeta function between these limits 2 and 1/2 is zero. We can reduce the zeta function to a second order differential equation which is seen to be elliptical in nature. Error Estimate Note Taking the McLauien series of the Zeta function in proper form, ie; f(0)= A0 + A0 . f’(0)/1! +A(0) f “(0)/2!+…….., we see that it grows at the rate of п Taking A0=1 the initial r of the Spiral observed by taking the polar form of the Zeta function we have, F1(0)=1+1. (dr/dt)/1! =2 F2(0)=1+2. (dr/dt)/1!=3 and so on. If we take trigonometric scale these become п, 2п, 3п and so on. which makes the error estimate as п x 1/ln(x) or п/ln (x). Here we see that the Zeta function grows at the rate of √2 or ln(п) as x grows as e√2=п, which is the proper form of en.log n the exponential growth which shows that zeta function values changes by 1.414 for each successive change in x which gives the zeta function it’s meaning.. The Spiral observed can be the spiral similar to the Planetary spiral of the Solar System or the Milky way. Conclusion Rieman’s Zeta function whose roots are ½ +/-0 remain proved. Reference: “God Created Integers” Steven Hawkings PP822 ______________ Mathew Cherian$$

Last edited: Apr 2, 2009
2. Apr 2, 2009

### CRGreathouse

Here's the first mistake I saw.

3. Apr 2, 2009

### imag94

$$You haven't noticed the transformation may be it is a translation errror, it should be read as, $$a= \ln (x)^x$$ The rational is since x is a unique value for any number say n we can take $$x^x$$ as a constant, then, $$a =\ln x^x = x. \ln x$$ (taking logrithm on both sides) I think it is a translation error and if you miss this transformation then the proof is difficult unless you go through other routes which I abstain here. One has to come through the reciprocal (for 1/ln x)of the expanded differential equation of the series expanstion of log function which is, $$\ln x = \frac {x-1}{x-2} + \frac {1}{3} \frac {x-3}{x-4} + \frac {1}{5} \frac {x-5}{x-6) .............$$ Again the expanded version should be integrated using Cauchy's principal number.$$

4. Apr 2, 2009

### AUMathTutor

Rationale has nothing to do with it.

a = x^x
...
a = x ln x

Can be true if and only if x^x = x ln x = ln (x^x). Coincidentally, this equation has *no* solutions at all.

Now let's pretend you meant

a = x^x
...
a = ln(x)^x

Here, x^x = ln(x)^x <=> x = ln(x) which is never true, either.

Sorry, I don't see it. The correct relationship would be:

a = x^x
ln a = x ln x.

5. Apr 2, 2009

### Hurkyl

Staff Emeritus
I don't think there's any point to this thread if the opening poster isn't even going to bother to get the typesetting correct!