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## how can we separate term of taylor series of k(x,t) (as a summation of product terms of fi(x) by gi(t))?

Asked by Alpa

### Alpa (view profile)

on 12 Jan 2013

Since any continuous function of two variables can be approximated by the Taylor expansion, then how this function can be separated as a summation of product terms of fi(x) by gi(t) (series)

for example: if ,then the Taylor expansion with respect the variable at with five terms is ,that mean ,and

now, I want (fi*gi) :

f1*g2=1*t=t
f1*g3=t2
f2*g3=x*t2

Azzi Abdelmalek

### Azzi Abdelmalek (view profile)

on 12 Jan 2013

Your expression is not clear

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## 1 Answer

### Roger Stafford (view profile)

Answer by Roger Stafford

### Roger Stafford (view profile)

on 12 Jan 2013
Edited by Roger Stafford

### Roger Stafford (view profile)

on 13 Jan 2013

Your statement "any continuous function of two variables can be approximated by the Taylor expansion" is not entirely correct. Only if the function is analytic in both variables is that true and then only within a certain region of analyticity. See

If f(x,y) can be expressed as a Taylor series about the point (0,0), its expansion would look like the following. Let f00 = f(0,0). Let fij be the i-th partial derivative with respect to x along with the j-th partial with respect to y evaluated at (0,0).

f(x,y) = (f00)/0! + (f10*x + f01*y)/1! + (f20*x^2+2*f11*x*y+f02*y^2)/2!
+ (f30*x^3+3*f21*x^2*y+3*f12*x*y^2+f03*y^3)/3!
+ (f40*x^4+4*f31*x^3*y+6*f22*x^2*y^2+4*f13*x*y^3+f04*y^4)/4!
+ (f50*x^5+5*f41*x^4*y+10*f32*x^3*y^2+10*f23*x^2*y^3+5*f14*x*y^4+f05*y^5)/5!
+ ...

where '!' denotes factorial. This gives you the first 21 terms and should show you the general pattern followed by subsequent terms. As you can see, the coefficients within the parentheses are the (i,j) binomial coefficients.

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