User:Eml4500.f08.delta 6.krueger/HW N1^i(x)tilda

From the diagram above it can be seen that $$\tilde{x}$$ is equal to $$x-x_{i}$$.Therefore the formula for $$N_{i}(\tilde{x})$$ from the above diagram is: $$N_{i}(\tilde{x})={x-x_{i} \over x_{i}-x_{i-1}}$$ It is also know that: $$N_{i}(\tilde{x})=N_{i}(x)=1-{x\over L^{(i)}}$$ Now let x=0 and x=$$L^{(i)}$$ and it can be seen that when: x=0 $$\Rightarrow$$ $$N_{i}(\tilde{x})=1$$ and when x=$$L^{(i)}$$ $$\Rightarrow$$ $$N_{i}(\tilde{x})=0$$