Approximation formula proof

[anhnha]

Hi.
Please help me prove the approximation formula below given in my book. This is not homework question.
Thanks.

 

[mfb]

You just neglect $\tau_1$ or $\tau_2$, whatever is smaller. Note that a larger value will lead to a larger exponential as well (for positive t).

 

[anhnha]

Thank you. However, that doesn't solve the problem. What need to be proved is different.

 

[Chestermiller]

$$\frac{τ_1e^{-\frac{t}{τ_1}}-τ_2e^{-\frac{t}{τ_2}}}{τ_1-τ_2}=\frac{e^{-\frac{t}{τ_1}}-(τ_2/τ_1)e^{-\frac{t}{τ_2}}}{1-(τ_2/τ_1)}=e^{-\frac{t}{τ_1}}\left(\frac{1-(τ_2/τ_1)e^{-t(\frac{1}{τ_2}-\frac{1}{τ_1})}}{1-(τ_2/τ_1)}\right)=e^{-\frac{t}{τ_1}}\left(\frac{1-(τ_2/τ_1)e^{-t\frac{(τ_1-τ_2)}{τ_1τ_2}}}{1-(τ_2/τ_1)}\right)$$

If τ₁>>τ₂, then $e^{-t\frac{(τ_1-τ_2)}{τ_1τ_2}}<1$.

From this, it follows that, in the numerator, $(τ_2/τ_1)e^{-t\frac{(τ_1-τ_2)}{τ_1τ_2}}<<1$

Also, in the denominator, $(τ_2/τ_1)<<1$

So the term in parenthesis approaches unity.

Chet

 

[mfb]

Thank you. However, that doesn't solve the problem. What need to be proved is different.

Why? It shows that the big ≈ is a correct approximation.

 

[Chestermiller]

I guess the analysis I did in #4 did not work for the OP? (It was just a more fleshed-out version of what mfb was saying).

Chet

 

[anhnha]

Thank you all.
I think you misread the question a bit. The expression on the right hand side of the equation is $$e^{-\frac{t}{τ}}$$ with $$τ = τ_1 + τ_2$$.

 

[Chestermiller]

Thank you all.
I think you misread the question a bit. The expression on the right hand side of the equation is $$e^{-\frac{t}{τ}}$$ with $$τ = τ_1 + τ_2$$.

You're right. But that doesn't matter much. The same general procedure could be used for this.

Chet