Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If somebody has a experience with polynomials.

How to set this Hermite function to have a general minimum where I want on $x$ axis, for example in $0.5$. Is it possible to be in analytic form $f[x,a]$? Point $a$ is minimum.

$$f[x] = -(1/48) + (3 x^2)/16 - (5 x^4)/16 + (7 x^6)/48$$

And the same thing with this Hermite function to have one of its roots where I want also in form $g[x,b]$? Point $b$ is one of zeros.

$$g[x] = -(x/2) + x^3/2$$

For Example I want to do this but without connecting two functions.

f[x] function is

 Plot[1/8 - x^2/4 + x^4/8, {x, -1, 1}]

and I moved the maximum in point a=47/133

Plot[Piecewise[{{2423393/2916000 + (831383 x)/972000 - (760627 x^2)/
 972000 - (2352637 x^3)/2916000, 
x < 47/133}, {-(17689/79507) + (2494149 x)/318028 - (2388015 x^2)/
 159014 + (2352637 x^3)/318028, x > 47/133}}], {x, -1, 1}]

Function g[x] is

Plot[-(x/2) + x^3/2, {x, -1, 1}]

and I moved zero in point b=47/133

Plot[Piecewise[{{47/180 - (43 x)/90 - (133 x^2)/180, 
x < 47/133}, {47/86 - (90 x)/43 + (133 x^2)/86, 
x > 47/133}}], {x, -1, 1}]

How to do this without connecting two functions?

share|improve this question

1 Answer 1

Your $f$ has a global minimum in $0$. Hence, if you want to build a function $f_1$ "with the same shape" of $f$ which attains its global minimum in $a$, then you should consider the function: $$f_1(x)=f(x-a)\; .$$

The same argument shows that: $$g_1(x)=g(x-b)$$ has one root in $b$, because $g$ has one root in $0$.

In general, you can reason as follows. Assume you want to find a function $f_1$ such that 1) its graph has "the same shape" of the graph of another known function $f$ and 2) it takes the value $f(x_0)$ in the point $x_1\neq x_0$ (this value could be a max, a min, a zero, or any arbitrary number in the range of $f$), i.e. $f_1(x_1)=f(x_0)$; then the function: $$f_1(x)=f(x-(x_1-x_0))$$ is the one you are looking for.

share|improve this answer
    
I didn't explain good, also the boundary conditions in -1 and 1 must be the same (my don't good explanation with "to have same shape"). Please, take a look up. I putted an example, but I don't want to connect two functions, I need just one. –  Pipe Dec 8 '11 at 1:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.